Question

A boy 11.0 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. If the boy throws the ball upward at 65.0 ∘ above the horizontal, at 9.50 m/s. How fast must the dog run to catch the ball just as it reaches the ground? How far from the tree will the dog catch the ball?

Answer 1

using the equation of linear motion we have

Y = Uyt - 0.5gt^2

-11 = 9.5sin 65 * t - 0.5*9.8*t^2

4.9t^2 - 8.61t - 11 = 0

t = 8.61 + sqrt(8.61^2 + 4*4.9*11) / 9.8

= 2.615 s

X = Ux*t = 9.5*cos 65 * 2.615 = 10.5 m

so the dog catches the ball 10.5 m from the tree

Vx = Ux = 9.5cos 65 = 4 m/s ( since there is no acceleration in the horizontal direction )

Vy = Uy - gt

= 9.5sin 65 - 9.8*2.615

= -17 m/s

Magnitude = sqrt(4^2 + 17^2 ) =17.5 m/s

direction = tan^-1 ( 17/4) = 76.76^o(below the horizontal)