\(\displaystyle \Delta y = V_i(t)+\frac{1}{2}at^2\)

Using the equation above and the given values, we can solve for the time. First we need to find the change in distance.

\(\displaystyle \Delta y=y_f-y_i=0m-3m=-3m\)

The plant travels \(\displaystyle \small -3\) meters in the downward direction. Now we can plug in the given values from the question to solve for time.

\(\displaystyle -3m= (0\frac{m}{s})+\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -3m = (-4.9\frac{m}{s^2})t^2\)

\(\displaystyle \frac {-3m}{-4.9\frac{m}{s^2}}=t^2\)

\(\displaystyle 0.612s^2=t^2\)

\(\displaystyle \sqrt{0.612s^2}=\sqrt{t^2}\)

\(\displaystyle 0.782s=t\)

The horizontal motion will not affect the time the disc is in the air. Time will be determined by the rate at which the disc falls: the acceleration due to gravity.

We can solve for the time using the equation below and the values given in the question.

\(\displaystyle \Delta y=V_i(t)+\frac{1}{2}at^2\)

\(\displaystyle (0m-1.5m)=(0\frac{m}{s})(t)+\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle (-1.5m)=\frac{1}{2}(-9.8\frac{m}{s^2})t^2\boldsymbol{}\)

\(\displaystyle (-1.5m)=(-4.9\frac{m}{s^2})t^2\)

\(\displaystyle \frac{-1.5m}{-4.9\frac{m}{s^2}}=t^2\)

\(\displaystyle 0.306s^2=t^2\)

\(\displaystyle \sqrt{0.306s^2}=\sqrt{t^2}\)

\(\displaystyle 0.553s=t\)

We are given the initial velocity, acceleration, and the initial and final distances.

\(\displaystyle \Delta y=y_f-y_i=0m-10m=-10m\)

Using the equation below and the given values, we can solve for the final velocity.

\(\displaystyle V_f^2=V_i^2+2a\Delta y\)

\(\displaystyle V_f^2=(0\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)\)

\(\displaystyle V_f^2=196\frac{m^2}{s^2}\)

\(\displaystyle \sqrt{V_f^2}=\sqrt{196\frac{m^2}{s^2}}\)

\(\displaystyle V_f=14\frac{m}{s}\)

The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

\(\displaystyle \Delta x=\frac{(v_f+v_i)}{2}\Delta t\)

Plug in our given values and solve.

\(\displaystyle 10m=\frac{(0\frac{m}{s}+v_i)}{2}5s\)

Divide both sides by \(\displaystyle \small 5\).

\(\displaystyle \frac{10m}{5s}=\frac{v_i}{2}\)

\(\displaystyle 2\frac{m}{s}=\frac{v_i}{2}\)

Multiply both sides by \(\displaystyle \small 2\).

\(\displaystyle 4\frac{m}{s}=v_i\)

All of the equations regarding acceleration require an initial velocity. We will need to solve for the initial velocity using the given variables. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

\(\displaystyle \Delta x=\frac{(v_f+v_i)}{2}\Delta t\)

Plug in our given values and solve.

\(\displaystyle 10m=\frac{(0\frac{m}{s}+v_i)}{2}5s\)

\(\displaystyle \frac{10m}{5s}=\frac{v_i}{2}\)

\(\displaystyle 2\frac{m}{s}=\frac{v_i}{2}\)

\(\displaystyle 4\frac{m}{s}=v_i\)

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

\(\displaystyle \Delta x=v_it+\frac{1}{2}at^2\)

Plug in the given values to solve for acceleration.

\(\displaystyle 10m=(4\frac{m}{s}) (5s)+\frac{1}{2}a(5s)^2\)

\(\displaystyle 10m=(20m)+\frac{1}{2}a(25s^2)\)

\(\displaystyle -10m=\frac{1}{2}a(25s^2)\)

\(\displaystyle -20m=a(25s^2)\)

\(\displaystyle \frac{-20m}{25s^2}=a\)

\(\displaystyle -0.8\frac{m}{s^2}=a\)

To solve for the final velocity, remember that the relationship between velocity, acceleration, and time is \(\displaystyle v_f=v_i+at\).

Using the given values for the initial velocity, acceleration, and time, we can solve for the final velocity.

\(\displaystyle v_f=1.5\frac{m}{s}+(2\frac{m}{s^2}*5.23s)\)

\(\displaystyle v_f=1.5\frac{m}{s}+10.46\frac{m}{s}\)

\(\displaystyle v_f=11.96\frac{m}{s}\)

We can solve this question using the equation for acceleration in terms of velocity:

\(\displaystyle a=\frac{v_2-v_1}{t}\)

We know our initial velocity (zero, since we start from rest), time, and the acceleration of gravity. Use these values to isolate the variable for the final velocity.

\(\displaystyle -9.8\frac{m}{s^2}=\frac{v_2-0\frac{m}{s}}{2.1s}\)

\(\displaystyle (-9.8\frac{m}{s^2})(2.1s)=v_2\)

\(\displaystyle -20.58\frac{m}{s}=v_2\)

Note that the final velocity is negative, since the book is traveling in the downward direction.

Use the kinematic equation:

\(\displaystyle v_{f} = v_{i} + at\) 

We are given the initial velocity, the time elapsed, and the acceleration. Using these values, we can solve for the final velocity.

\(\displaystyle v_f=25\frac{m}{s}+(2\frac{m}{s^2})(15s)\)

\(\displaystyle v_f=25\frac{m}{s}+30\frac{m}{s}\)

\(\displaystyle v_f=55\frac{m}{s}\)

Utilize the kinematic equation:

\(\displaystyle x = v_{o}t + \frac{1}{2}at^{2}\)

The particle's motion can be broken into two parts: the initial distance and the distance traveled during acceleration. The initial distance is given.

\(\displaystyle x_o=35000m\)

The distance during acceleration can be found using the kinematic formula and given values for the initial velocity, acceleration, and time.

\(\displaystyle x_a = (850\frac{m}{s})(15s) + \frac{1}{2}(12\frac{m}{s^2})(15s)^{2}\)

\(\displaystyle x_a=(12750m)+(1350m)\)

\(\displaystyle x_a=14100m\)

Add the two distances together.

\(\displaystyle x_o+x_a=35000m+14100m=49100m\)

Convert the final answer to kilometers.

\(\displaystyle 49100m*\frac{1km}{1000m}=49.1km\)

Although this question involves several steps, when you break it down we can see that it is a problem involving kinematics in two dimensions.

First, begin by writing down what we know:

\(\displaystyle t=3.45s\)

Distance to goal post: \(\displaystyle x_1=58.5m\)

Distance past goal post: \(\displaystyle x_2=4.50m\)

What we want to know:

\(\displaystyle v_0=?\)

\(\displaystyle \theta=?\)

To begin, we need to calculate each of the vector components of the velocity. We can start with the horizontal component. We know that velocity is equal to distance divided by time, and that horizontal velocity, \(\displaystyle v_x\), will not change because there is no acceleration in the horizontal direction. We will need to find the total distance that the ball travels in order to solve.

\(\displaystyle v_x=\frac{x}{t}\)

We will need to find the total distance that the ball travels in order to solve.

\(\displaystyle x=x_1+x_2\)

\(\displaystyle x=58.5m+4.50m\)

\(\displaystyle x=63.0m\)

Use this distance and the given time to find the horizontal velocity.

\(\displaystyle v_x=\frac{63.0m}{3.45s}\)

\(\displaystyle v_x=18.3\frac{m}{s}\)

Now let's find the initial vertical velocity,\(\displaystyle v_{oy}\). Because we are assuming that there is nothing except for gravity influencing the ball, we can say that the ball spends half of the time reaching the peak of its trajectory, where the vertical velocity will momentarily be zero. With that information, we can solve for the initial vertical velocity:

\(\displaystyle v_y=v_{oy}+a(\frac{1}{2}t)\)

We use only half the given time because we are only taking the time from when the ball is kicked to when it reaches the top of its trajectory (which will be half of its flight). As we stated above, velocity will be zero at the top of the trajectory. We are using a negative value for the acceleration due to gravity because gravity points downward, which in this case is the negative direction. Use the given values to solve for the initial vertical velocity.

\(\displaystyle v_{oy}=v_y-a(\frac{1}{2}t)\)

\(\displaystyle v_{oy}=0\frac{m}{s}-(-9.81\frac{m}{s^2})(\frac{1}{2}*3.45s)\)

\(\displaystyle v_{oy}=16.9\frac{m}{s}\)

Now that we have both directional components of the initial velocity, we can use the Pythagorean theorem to solve for the total initial velocity.

\(\displaystyle {v_o}^2={v_x}^2+{v_o_y}^2\)

\(\displaystyle v_o=\sqrt{{v_x}^2+{v_o_y}^2}\)

\(\displaystyle v_o=\sqrt{(18.3\frac{m}{s})^2+(16.9\frac{m}{s})^2}\)

\(\displaystyle v_o=24.9\frac{m}{s}\)

Now to find the angle, we use trigonometry. In a triangle formed by the maximum height of the ball, the ground, and the trajectory, tangent of the angle will be equal to the vertical leg of the triangle divided by the horizontal leg of the triangle.

\(\displaystyle tan\theta=\frac{v_{oy}}{v_x}\)

\(\displaystyle \theta=tan^{-1}(\frac{v_{oy}}{v_x})\)

\(\displaystyle \theta=tan^{-1}(\frac{16.9}{18.3})\)

\(\displaystyle \theta=42.8^0\)