(Solved): (Figure 1)A cannonball is fired horizontally from the top of a cliff. The cannon is at height \( H= ...

(Figure 1)A cannonball is fired horizontally from the top of a cliff. The cannon is at height \( H=100 \mathrm{~m} \) above ground level, and the ball is fired with initial horizontal speed \( v_{0} \). Assume acceleration due to gravity to be \( g=9.80 \mathrm{~m} / \mathrm{s}^{2} \). Assume that the cannon is fired at time \( t=0 \) and that the cannonball hits the ground at time \( t_{\mathrm{g}} \). What is the \( y \) position of the cannonball at the time \( t_{g} / 2 \) ? Answer numerically in units of meters. View Available Hint(s) Correct The same answer can be obtained more easily (perhaps you did it this way) if you notice that \( v_{0 y}=0 \). This means that the vertical displacement is given by \( \Delta y=-\frac{1}{2} g t^{2} \) and therefore \( \Delta y\left(t_{\mathrm{g}} / 2\right) \) is one-quarter of \( H \); then \( y\left(t_{\mathrm{g}} / 2\right)=\frac{3}{4} H \). Part B Given that the projectile lands at a distance \( D=200 \mathrm{~m} \) from the cliff, as shown in the figure, find the initial speed of the projectile, \( v_{0} \). Express the initial speed numerically in meters per second. View Available Hint(s) What is the \( y \) position of the cannonball when it is at distance \( D / 2 \) from the hill? If you need to, you can use the trajectory equation for this projectile, which gives \( y \) in terms of \( x \) directly: \[ y=H-\frac{g x^{2}}{2 v_{0 x}^{2}} . \] You should already know \( v_{0 x} \) from the previous part. Express the position of the cannonball numerically in meters.