My Notes Ask Your Teach A woman throws a ball at a vertical walld-6.4 m away. Th

Question

My Notes Ask Your Teach A woman throws a ball at a vertical walld-6.4 m away. The ball is h 32 m above ground when the horizontal component of its velocity the vertical ground when it leaves the woman's hand with an initial velocity of 15 mys at 45 When the ball hits the wall, leaves is reversed, the vertical component remains unchanged. (ignore any effects due to air resistance.) (a) Where does the ball hit the ground? m (away from the wall) (b) How long was the ball in the air before it hit the wall? (c) Where did the ball hit the wall? m (above the ground) (d) How long was the ball in the air after it left the wall?

Explanation / Answer

from hand of woman to wall,

in horizontal, d = (v0x) t

6.4 = (15 cos45) t

t = 0.603 sec

v0x = 15 cos45 = 10.6 m/s

in vertical, yf - 3.2 = (15 sin45)(0.603) - (9.8 x 0.603^2)/2

yf = 7.8 m

vy = (15 sin45) - (9.8 x 0.603) = 4.70 m/s

after that from wall to ground,

yf - yi = vy t + ay t^2 / 2

0 - 7.8 = 4.70 t - 9.8 t^2 / 2

4.9 t^2 - 4.70 t - 7.8 = 0

t = 1.83 sec

(A) distance from wall = (15 cos45) (1.83)

= 19.4 m

(B) t = 0.6602 sec

(C) yf = 7.8 m

(D) t = 1.83 sec

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