{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

90_Physics ProblemsTechnical Physics

# 90_Physics ProblemsTechnical Physics - Chapter 4*P4.21(a...

This preview shows page 1. Sign up to view the full content.

Chapter 4 91 *P4.21 (a) For the horizontal motion, we have x x v t a t v v f i xi x i i = + + = + ° + = 1 2 24 0 53 2 2 0 18 1 2 m s m s cos . . . a fa f (b) As it passes over the wall, the ball is above the street by y y v t a t f i yi y = + + 1 2 2 y f = + ° + = 0 18 1 53 2 2 1 2 9 8 2 2 8 13 2 . sin . . . . m s s m s s m 2 b ga fa f e j a f . So it clears the parapet by 8 13 7 1 13 . . m m m = . (c) Note that the highest point of the ball’s trajectory is not directly above the wall. For the whole flight, we have from the trajectory equation y x g v x f i f i i f = F H G I K J tan cos θ θ b g 2 2 2 2 or 6 53 9 8 2 18 1 53 2 2 2 m m s m s 2 = ° ° F H G G I K
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online