## College Physics (4th Edition)

(a) The ball strikes the wall at a height of $1.10~m$ above the ground. (b) The ball is on its way down when it hits the wall.
(a) We can find the time it takes to travel a horizontal distance of $10~m$: $t = \frac{10~m}{(20~m/s)~sin~80^{\circ}} = 0.508~s$ We can find the height of the ball at this time: $y = y_0+v_{0y}t+\frac{1}{2}a_yt^2$ $y = 0.60~m+(20~m/s)~cos~80^{\circ}~(0.508~s)+\frac{1}{2}(-9.80~m/s^2)(0.508~s)^2$ $y = 1.10~m$ The ball strikes the wall at a height of $1.10~m$ above the ground. (b) We can find the vertical component of the velocity at $t = 0.508~s$: $v_{fy} = v_{0y}+a_y~t$ $v_{fy} = (20~m/s)~cos~80^{\circ}+(-9.80~m/s^2)(0.508~s)$ $v_{fy} = -1.51~m/s$ The negative value for the vertical component of the velocity shows that the ball is on its way down when it hits the wall.