Question

In the figure, a baseball is hit at a height h = 1.20 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.1 s after it is hit and then down past the top of the wall 4.8 s later, at distance D = 54 m farther along the wall. (a) What horizontal distance is traveled by the ball from hit to catch? What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's velocity just after being hit? (d) How high is the wall?

Answer 1

Answer:

D.

Explanation:

The ball travels distance D in ∆t = 2.60 s. Therefore, the x component of the initial velocity is D =v

ox

​

 ⇒ ∆t = D/∆t = 15.4 m/s Due to the symmetry of projectile motions, it takes 1.2 seconds the ball to reach the initial height after passing the top of the wall downward. Therefore, the total flight time is t

tot

​

 = 1.2 + 2.6 +1.2 = 5 s The horizontal range is R = v

ox

​

t

tot

​

 = 77.0 m  

By symmetry, the ball reaches the peak of the motion at time t

top

​

 = t

tot

​

/2 = 2.5 s At the peak, the y component of the velocity is zero. Thus we have 0 = v

oy

​

 – gt

top

​

 ⇒ v

oy

​

  = gt

top

​

= 24.5 m/s The height of the wall is equal to the height of the ball at t = 1.2 s h = y (t = 1.2 s) = v

oy

​

 +v

oy

​

t – (1/2)gt

2

  = 23.3 m