A boy 12.0 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 horizontally at 8.50 m/s, (a) how fast must the dog run to catch the ball just as it reaches the ground, and (b) how far from the tree will the dog catch the ball?

Question

Averi

Physics

26 April, 17:00

A boy 12.0 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 horizontally at 8.50 m/s, (a) how fast must the dog run to catch the ball just as it reaches the ground, and (b) how far from the tree will the dog catch the ball?

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Answers (1)

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Gianni Cardenas · Answer 1

Height covered = 12m

time to fall by 12 m

s = 1/2 gt²

12 = 1/2 g t²

t = 1.565 s

Horizontal distance of throw

= 8.5 x 1.565

= 13.3 m

This distance is to be covered by dog during the time ball falls ie 1.565 s

Speed of dog required = 13.3 / 1.565

= 8.5 m / s

b) dog will catch the ball at a distance of 13.3 m.