The key to solving this problem is to recognize that in two dimensional motion, the x-direction and the y-direction are dealt with separately. The question asks how long the projectile is in the air so we are dealing only with motion in the y-axis. 

Since the projectile was fired at an angle we must use trigonometry to find the velocity in the y-axis \(\displaystyle (V_y)\). Then we use a kinematic equation to solve for time.

Initial velocity in the y-axis:

\(\displaystyle V_y=200*\sin30^{\circ}=100\frac{m}{s}\)

Now we will find the time it takes for the projectile to hit the ground using a kinematic equation:

\(\displaystyle V_y=V_{0y}+a_yt\)

We use \(\displaystyle V_y=0\)at the top of the projectile's trajectory, where its velocity is changing from going up, to going down. 

\(\displaystyle 0=100-10t\)

\(\displaystyle 10t=100\)

\(\displaystyle t=10s\)

Note that this is the time it takes for the projectile to reach maximum height. The time it takes to get back to the ground from here is the same as the time it took to get here. Thus, to find the total flight time, we must double our answer.

\(\displaystyle t_{total}=20s\)

The question asks how far the projectile can travel before landing. Since we deal with the x and y-axes of 2-dimensional motion separately, it is important to recognize that only the movement in the x-axis is relevant. If you are driving 50mph and do so for three hours you have driven 150mi. Similarly, our projectile is traveling at a certain speed horizontally and does so for 20s. Thus we simply multiply its horizontal speed by the time the projectile was in the air. Since the projectile was launched at an angle we use trigonometry to find its horizontal speed:

\(\displaystyle V_x= 200*cos(30^{\circ})=173\frac{m}{s}\)

Now use the formula for range and plug in the velocity in the x-direction (constant) to find the distance travelled.

\(\displaystyle \Delta x=V_x*t=173\frac{m}{s}*20s=3460m\)

Acceleration is defined by the rate of change of velocity per unit time. Thus, if the acceleration of gravity on Earth is \(\displaystyle -10 \frac{m}{s^2}\) (which is an approximation close enough to the actual value for our purposes) then all objects are accelerated downward by 10 meters per second every second. Since the ball was thrown directly upward and will fall straight down, we need not worry about sideways motion and can focus on gravity slowing the ball until at reaches a maximum height and falls back to Earth. The ball was thrown upward with an initial velocity of \(\displaystyle 22\frac{m}s{}\). Gravity will slow the ball down by \(\displaystyle 10 \frac{m}{s}\) until it reaches a maximum height. Thus, the ball will reach this maximum height in 2.2s. Mathematically: \(\displaystyle \frac{22\frac{m}{s}}{10\frac{m}{s^2}}= 2.2 s\)

It then falls back to Earth for 2.2s. Thus, the ball takes a total of 4.4s to reach the ground.

The question gives the total initial velocity, but we will need to find the horizontal component using trigonometry.

To find the horizontal velocity we use the equation \(\displaystyle v_{i}*\cos(\Theta)=v_{x}\).

We can plug in the given values for the angle and initial velocity to solve.

\(\displaystyle 30\frac{m}{s}*\cos(35^{\circ})=v_{x}\)

\(\displaystyle 30\frac{m}{s}*(0.82)=v_{x}\)

\(\displaystyle 24.57\frac{m}{s}=v_{x}\)

Even though the problem gives us an initial velocity, we need to break it down into horizontal and vertical components.

\(\displaystyle v_{i} *\sin(\Theta)=v_{iy}\)

We can plug in the given values and find the vertical velocity.

\(\displaystyle 30\frac{m}{s}* \sin(35^{\circ})=v_{iy}\)

\(\displaystyle 30\frac{m}{s}* (0.57)=v_{iy}\)

\(\displaystyle 17.21\frac{m}{s}=v_{iy}\)

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial and final vertical velocities, we can plug our values into an equation to solve for the maximum height.

\(\displaystyle v_{f}^2=v_i^2+2a\Delta y\)

\(\displaystyle (0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y\)

\(\displaystyle (0\frac{m}{s})=(296.09\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y\)

\(\displaystyle -(296.09\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y\)

\(\displaystyle \frac{-296.09\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y\)

\(\displaystyle 15.11m=\Delta y\)

Remember, \(\displaystyle \Delta y\) only tells us the CHANGE in the vertical direction. The rock started at the top of a \(\displaystyle 10m\) tall building, then rose an extra \(\displaystyle 15.11m\).

\(\displaystyle 10m+15.11m=25.11m\)

Its highest point is \(\displaystyle 25.11m\) above the ground.

We first need to find the vertical component of the velocity.

\(\displaystyle v_{i}*\sin(\Theta)=v_{iy}\)

We can plug in the given values for the angle and the initial velocity to find the vertical component.

\(\displaystyle 30\frac{m}{s}* \sin(35^{\circ})=v_{iy}\)

\(\displaystyle 30\frac{m}{s}* (0.57)=v_{iy}\)

\(\displaystyle 17.21\frac{m}{s}=v_{iy}\)

Now we need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air.

Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward.

\(\displaystyle v_f=v_i+at_1\)

\(\displaystyle 0\frac{m}{s}=17.21\frac{m}{s}+(-9.8\frac{m}{s^2})t_1\)

\(\displaystyle -17.21\frac{m}{s}=(-9.8\frac{m}{s^2}) t_1\)

\(\displaystyle \frac{-17.21\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1\)

\(\displaystyle 1.76s=t_1\)

Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward.

\(\displaystyle v_f^2=v_i^2+2a\Delta y\)

\(\displaystyle (0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y\)

\(\displaystyle (0\frac{m}{s})=(296.18\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y\)

\(\displaystyle -(296.18\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y\)

\(\displaystyle \frac{-296.18\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y\)

\(\displaystyle 15.11m=\Delta y\)

Remember, \(\displaystyle \Delta y\) only tells us the vertical CHANGE. Since the rock started at the top of a \(\displaystyle 10m\) building, if it rose an extra \(\displaystyle 15.11m\), then at its highest point it is \(\displaystyle 25.11m\) above the ground.

This means that our \(\displaystyle \Delta y\) will be \(\displaystyle -25.11m\) as it will be traveling down from the highest point. Using this distance, we can find the downward travel time.

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

\(\displaystyle -25.11m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -25.11m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -25.11m=(-4.9\frac{m}{s^2})t^2\)

\(\displaystyle \frac{-25.11m}{-4.9\frac{m}{s^2}}=t^2\)

\(\displaystyle 5.12s^2=t^2\)

\(\displaystyle \sqrt{5.12s^2}=\sqrt{t^2}\)

\(\displaystyle 2.26s=t_2\)

Add together the time for upward travel and downward travel to find the total flight time.

\(\displaystyle t_{total}=t_1+t_2\)

\(\displaystyle t_{total}=1.76s+2.26s\)

\(\displaystyle t_{total}=4.02s\)

We first need to find the horizontal component of the initial velocity.

\(\displaystyle v_{i}* \cos(\Theta)=v_{x}\)

We can plug in the given values for the angle and initial velocity and solve.

\(\displaystyle 30\frac{m}{s}* \cos(35^{\circ})=v_{x}\)

\(\displaystyle 30\frac{m}{s}* (0.82)=v_{x}\)

\(\displaystyle 24.57\frac{m}{s}=v_{x}\)

The only force acting on the rock during flight is gravity; there are no forces in the horizontal direction, meaning that the horizontal velocity will remain constant. We can set up a simple equation to find the relationship between distance traveled and the velocity.

\(\displaystyle \Delta x =v_xt\)

We know \(\displaystyle v_x\), but now we need to find the time the rock is in the air.

We need to solve for the time that the rock travels upward. We can then add the upward travel time to the downward travel time to find the total time in the air.

Remember that the vertical velocity at the highest point of a parabola is zero. We can use that to find the time for the rock to travel upward.

\(\displaystyle v_f=v_i+at_1\)

\(\displaystyle 0\frac{m}{s}=17.21\frac{m}{s}+(-9.8\frac{m}{s^2})t_1\)

\(\displaystyle -17.21\frac{m}{s}=(-9.8\frac{m}{s^2}) t_1\)

\(\displaystyle \frac{-17.21\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1\)

\(\displaystyle 1.76s=t_1\)

Now let's find the time for the downward travel. We don't know the final velocity for the rock, but we CAN use the information we have been given to find the height it travels upward.

\(\displaystyle v_f^2=v_i^2+2a\Delta y\)

\(\displaystyle (0\frac{m}{s})^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y\)

\(\displaystyle (0\frac{m}{s})=(296.18\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y\)

\(\displaystyle -(296.18\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y\)

\(\displaystyle \frac{-296.18\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y\)

\(\displaystyle 15.11m=\Delta y\)

Remember, \(\displaystyle \Delta y\) only tells us the vertical CHANGE. Since the rock started at the top of a \(\displaystyle 10m\) building, if it rose an extra \(\displaystyle 15.11m\), then at its highest point it is \(\displaystyle 25.11m\) above the ground.

This means that our \(\displaystyle \Delta y\) will be \(\displaystyle -25.11m\) as it will be traveling down from the highest point. Using this distance, we can find the downward travel time.

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

\(\displaystyle -25.11m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -25.11m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2\)

\(\displaystyle -25.11m=(-4.9\frac{m}{s^2})t^2\)

\(\displaystyle \frac{-25.11m}{-4.9\frac{m}{s^2}}=t^2\)

\(\displaystyle 5.12s^2=t^2\)

\(\displaystyle \sqrt{5.12s^2}=\sqrt{t^2}\)

\(\displaystyle 2.26s=t_2\)

Add together the time for upward travel and downward travel to find the total flight time.

\(\displaystyle t_{total}=t_1+t_2\)

\(\displaystyle t_{total}=1.76s+2.26s\)

\(\displaystyle t_{total}=4.02s\)

Now that we've finally found our time, we can plug that back into the equation from the beginning of the problem, along with our horizontal velocity, to solve for the final distance.

\(\displaystyle \Delta x =v_xt\)

\(\displaystyle \Delta x =24.57\frac{m}{s}* 4.02s\)

\(\displaystyle \Delta x=98.77m\)

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation \(\displaystyle v_{i}*\cos(\Theta)=v_{x}\).

We can plug in the given values for the angle and initial velocity to solve.

\(\displaystyle 30\frac{m}{s}*\cos(35^{\circ})=v_{x}\)

\(\displaystyle 30\frac{m}{s}*(0.82)=v_{x}\)

\(\displaystyle 24.57\frac{m}{s}=v_{x}\)

We can find the vertical velocity using the equation \(\displaystyle v_{i}* \sin(\Theta)=v_{iy}\).

\(\displaystyle 30\frac{m}{s}* \sin(35^{\circ})=v_{iy}\)

\(\displaystyle 30\frac{m}{s}* (0.57)=v_{iy}\)

\(\displaystyle 17.21\frac{m}{s}=v_{iy}\)

The horizontal velocity will not change during flight because there are no forces in the horizontal direction. The vertical velocity, however, will be affected. We need to solve for the final vertical velocity, then combine the vertical and horizontal vectors to find the total final velocity.

We know that the rock is going to travel a net distance of \(\displaystyle -10m\), as that is the distance between where the rock's initial and final positions. We now know the displacement, initial velocity, and acceleration, which will allow us to solve for the final velocity.

\(\displaystyle v_f^2=v_1^2+2a\Delta y\)

\(\displaystyle v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)\)

\(\displaystyle v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}\)

\(\displaystyle v_f^2=492.18\frac{m^2}{s^2}\)

\(\displaystyle \sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}\)

\(\displaystyle v_f=22.19\frac{m}{s}\)

Because the rock is traveling downward, our velocity will be negative: \(\displaystyle v_f=-22.19\frac{m}{s}\).

Now that we know our final velocities in both the horizontal and vertical directions, we can use the Pythagorean theorem to solve for the net velocity.

\(\displaystyle (-22.19\frac{m}{s})^2+(24.57\frac{m}{s})^2=(v_{resultant})^2\)

\(\displaystyle 492.4\frac{m^2}{s^2}+603.68\frac{m^2}{s^2}=(v_{resultant})^2\)

\(\displaystyle 1096.08\frac{m^2}{s^2}=(v_{resultant})^2\)

\(\displaystyle \sqrt{1096.08\frac{m^2}{s^2}}=\sqrt{(v_{resultant})^2}\)

\(\displaystyle 33.1\frac{m}{s}=v_{resultant}\)

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation \(\displaystyle v_{i}*\cos(\Theta)=v_{x}\).

We can plug in the given values for the angle and initial velocity to solve.

\(\displaystyle 30\frac{m}{s}*\cos(35^{\circ})=v_{x}\)

\(\displaystyle 30\frac{m}{s}*(0.82)=v_{x}\)

\(\displaystyle 24.57\frac{m}{s}=v_{x}\)

We can find the vertical velocity using the equation \(\displaystyle v_{i}* \sin(\Theta)=v_{iy}\).

\(\displaystyle 30\frac{m}{s}* \sin(35^{\circ})=v_{iy}\)

\(\displaystyle 30\frac{m}{s}* (0.57)=v_{iy}\)

\(\displaystyle 17.21\frac{m}{s}=v_{iy}\)

The horizontal velocity will not change during flight because there are no forces in the horizontal direction. The vertical velocity, however, will be affected. We need to solve for the final vertical velocity, then combine the vertical and horizontal vectors to find the total final velocity.

We know that the rock is going to travel a net distance of \(\displaystyle -10m\), as that is the distance between where the rock's initial and final positions. We now know the displacement, initial velocity, and acceleration, which will allow us to solve for the final velocity.

\(\displaystyle v_f^2=v_1^2+2a\Delta y\)

\(\displaystyle v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)\)

\(\displaystyle v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}\)

\(\displaystyle v_f^2=492.18\frac{m^2}{s^2}\)

\(\displaystyle \sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}\)

\(\displaystyle v_f=22.19\frac{m}{s}\)

Because the rock is traveling downward, our velocity will be negative: \(\displaystyle v_f=-22.19\frac{m}{s}\).

Now that we know our final velocities in both the horizontal and vertical directions, we can find the angle created between the two trajectories. The horizontal and vertical velocities can be compared using trigonometry.

\(\displaystyle \tan(\Theta)=\frac{v_y}{v_x}\),

\(\displaystyle \tan^{-1}(\frac{v_y}{v_x})=\Theta\)

Plug in our values and solve for the angle.

\(\displaystyle \tan^{-1}(\frac{22.19\frac{m}{s}}{24.57\frac{m}{s}})=\Theta\)

\(\displaystyle \tan^{-1}(0.90)=\Theta\)

\(\displaystyle 41.99^{\circ}=\Theta\)

The question gives the total initial velocity, but we will need to find the horizontal and vertical components.

To find the horizontal velocity we use the equation \(\displaystyle v_{i}*\cos(\Theta)=v_{x}\).

We can plug in the given values for the angle and initial velocity to solve.

\(\displaystyle 30\frac{m}{s}*\cos(35^{\circ})=v_{x}\)

\(\displaystyle 30\frac{m}{s}*(0.82)=v_{x}\)

\(\displaystyle 24.57\frac{m}{s}=v_{x}\)

We can find the vertical velocity using the equation \(\displaystyle v_{i}* \sin(\Theta)=v_{iy}\).

\(\displaystyle 30\frac{m}{s}* \sin(35^{\circ})=v_{iy}\)

\(\displaystyle 30\frac{m}{s}* (0.57)=v_{iy}\)

\(\displaystyle 17.21\frac{m}{s}=v_{iy}\)

The question asks for the total velocity at the maximum height. At the top of the parabola, \(\displaystyle v_y=0\frac{m}{s}\) as the direction of the motion changes from upward to downward. Yet, even though it has no vertical velocity, the horizontal velocity remains constant during flight. The only force on the rock is that of gravity, and gravity will only affect the vertical velocity. There are no horizontal forces present to alter the horizontal velocity.

Our final answer will be equal to the horizontal velocity: \(\displaystyle 24.57\frac{m}{s}\).