Abatter hits a pitched ball when the center of the ball is 1.22 m above the ground. the ball leaves the bat at an angle of 45° with the ground. with that launch, the ball should have a horizontal range (returning to the launch level) of 107 m.
(a) does the ball clear a 7.32-m-high fence that is 97.5 m horizontally from the launch point?
(b) at the fence, what is the distance between the fence top and the ball center?

2.56m

Explanation:

This is problem 4.47 from your textbook.

Find: Whether a batted baseball clears a fence, and by what amount it does or does not.

Given: The baseball’s initial launch height and angle, the range the baseball would have without the fence, the distance to the fence and its height.

Let the y axis run vertically and the x axis horizontally. Let the range the baseball would have without the fence be R=107 m, with the distance to the fence d=97.5m and its height hfence=7.32 m. The baseball is batted at an angle θ=45° at speed vi a height of hbat=1.22m above the ground.

Let the origin be at the position the ball leaves the bat. The height of the fence relative to the height of the bat is then

δh = hfence − hbat

What we really need to determine is the ball’s y coordinate at x = d. If y > δh, the ball clears the fence. We can use the range the baseball would have without the fence and the launch angle to find the ball’s speed, which will allow a complete calculation of the trajectory.

Relevant equations: We need only the equations for the range and trajectory of a projectile over level ground:

R = (vi*sin2θ)/g

For convenience sake and easy reading, I extracted the solution of your textbook for the remaining parts of the solution.

Hence, it is seen that the ball does clear the fence, by approximately 2.56 m

you have to draw the persons position over there