Question

A boy 13.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the instant the ball is thrown. (a) If the boy throws the ball horizontally at 7.50 m/s, how fast must the dog run to catch the ball just as it reaches the ground? (b) How far from the tree will the dog catch the ball? (c) Find the velocity of the ball (magnitude and direction) just as it reaches the ground.

Answer 1

d = 13 m , u = 7.5 m/s

(a) There is no acceleration in the x-direction, so the ball will have a constant horizontal component of velocity. If the dog is going to catch the ball, his horizontal velocity must be the same as the ball (the ball will be directly above the dog the whole time).

v = 7.5 m/s

(b)

from kinematic eqaution

y = yo +uyt+(1/2)at^2

0 = 13 +0 - 0.5*9.8*t^2

t = 1.63 s

x = ux*t

x = 7.5*1.63 =12.225 m

(c) from kinematic eqautions

v^2 - u^2 = 2as

vy^2 = 2*9.8*13

vy = 15.96 m/s

vx = 7.5 m/s

v = (7.5^2 +15.96^2)^0.5

v = 17.63 m/s

tan(theta) = 15.96/7.5

theta= 64.83 degrees