a. 1.51 s
In this part of the problem, we are only interested in the horizontal motion of the ball. Along the horizontal direction, the motion of the ball is a uniform motion with constant velocity, which is equal to the horizontal component of the initial velocity:
[tex]v_x = v_0 cos \theta = (19.0 m/s)(cos 40^{\circ})=14.6 m/s[/tex]
So, the ball travels horizontally at a speed of 14.6 m/s; in order to cover the distance of d = 22.0 m that separates it from the wall, the time need is
[tex]t=\frac{d}{v_x}=\frac{22.0 m}{14.6 m/s}=1.51 s[/tex]
b. 7.25 m
We now know that the ball takes 1.51 s to hit the wall, 22.0 away. Now we have to analyze the vertical motion of the ball, which is an accelerated motion with constant acceleration g =9.8 m/s^2 towards the ground (acceleration due to gravity).
The initial vertical velocity is
[tex]v_{y0}=v_0 sin \theta=(19.0 m/s)(sin 40^{\circ})=12.2 m/s[/tex]
The vertical position of the ball at time t is given by the equation
[tex]y(t)=v_{0y} t -\frac{1}{2}gt^2[/tex]
We know that the ball hits the wall at t=1.51 s, so if we substitute this value into the previous formula, we find at what height y the ball hits the wall:
[tex]y(1.51 s)=(12.2 m/s)(1.51 s)-\frac{1}{2}(9.8 m/s^2)(1.51 s)^2=7.25 m[/tex]
