Calculating Motion in Two Dimensions - AP Physics C: Electricity and Magnetism

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Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

Compare your answer with the correct one above

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