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 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.

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

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

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

We are looking for total velocity, which in this case has both a horizontal and vertical component.

Because the helicopter is flying horizontally we know \(\displaystyle v_{y0}=0\)

We can assume that this takes place near the surface of the earth so \(\displaystyle a=g\)

We can plug this into the equation:

\(\displaystyle v_{yf}=v_{y0} +at\)

\(\displaystyle v_{yf}=0 +(10)(8)\)

\(\displaystyle v_{yf}=80\)

Next we must find the horizontal velocity. Because there are no additional forces, the horizontal velocity is the same as the initial horizontal velocity of the helicopter, so:

\(\displaystyle v_{x}=60\)

Next we must use vector addition to add the horizontal and vertical components of velocity. Because this horizontal and vertical velocities are perpendicular the sum will be the hypotenuse of a right triangle:

\(\displaystyle 60^{2}+80^{2}=v^{2}\)

\(\displaystyle v=100\frac{m}{s}\)