Need Physics help! Please show all work in clear writing! .55 In Fig. 4-48, a ba

Question

Need Physics help! Please show all work in clear writing!

.55 In Fig. 4-48, a baseball is hit at a height h- 1.00 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.00 s after it is hit and then down past the top of the wall 4.00 s later, at distance D - 50.0 m farther along the wall. (a) What horizontal distance is trav- eled 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? FIG. 4-48 Problem 55.

Explanation / Answer

Since ball travels 50 m in 4 s

Ux = 50/4 = 12.5m/s

Since ball reaches the top height in 3 sec, (because 1 sec to cross the wall and half of 4s to reach max height)

Vy= Uy + at

0 = Uy -9.8 x 3
Uy = 29.4 m/s

a) by symmetry, it took 6 sec from hit to catch

x = 6 x Ux = 75 m

b) U^2 = Ux^2 + Uy^2 = 12.5^2 + 29.4^2

U = 31.95 m/s

c)

@ = tan-1(Uy/Ux) = 67 degree

d) using second kinematical equation

s = Uy t + 1/2 at^2

= 29.4 x 1 -1/2 x 9.8 x1^2

= 24.5 m

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