A woman throws a ball at a vertical wall d 4.6 m away. The ball is h 2.3 m above

Question

A woman throws a ball at a vertical wall d 4.6 m away. The ball is h 2.3 m above ground when it leaves the woman's hand with an initial velocity of 10 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity 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? 0.65 (c) Where did the ball hit the wall? 4.82 m (above the ground) (d) How long was the ball in the air after it left the wall?

Explanation / Answer

Given,

d = 4.6 m ; h = 2.3 m ; v0 = 10 m/s ; theta = 45

v0x = 10 cos45 = 7.07 m/s

v0y = 10 sin45 = 7.07 m/s

we know that,

R = 4.6/7.07 = 0.65 m

H = 2.3 + 7.07 x 0.65 - 0.5 x 9.8 x 0.65^2 = 4.83 m

a)R = vx T

R = 7.07 x 0.99 = 6.99 m

Hence, R = 6.99 m = 7 m

(T calculated in later part)

d)T = sqrt (2H/g)

T = sqrt (2 x 4.83/9.8) = 0.99 m = 1 m

Hence, T = 0.99m = 1 s

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