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3 years ago

7

Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the gr

ound. When the balls collide, they are moving n opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur? (Use any variable or symbol stated above as necessary.) Both balls experience constant acceleration once they are �n flight. Choose a coordinate system with the origin at the ground and the upward direction positive. When the balls collide they are at the same height above the ground.

1 answer:

kramer3 years ago

3 0

Answer:

2h / 3

Explanation:

Let after time t , collision occurs

distance covered by A

= 1/2gt²

Let u be the initial velocity with which B was thrown upwards

distance by B

= ut - 1/2 gt²

total distance by A and B

h = ut - 1/2 gt² + 1/2gt²

h = ut

t = h / u

velocity of A after t

= gt

= g x h/u

velocity of B after t

= u - gt

= u - g x h/u

given

g x h/u = 2 (u - g x h/u)

2u² - 2gh = gh

u = $\sqrt{\frac{3gh}{2} }$

t = h / u

= $\sqrt{\frac{2h}{3g} }$

Distance covered by A

= 1/2 gt²

= 1/2 x 2h/3

= h/3

Height at which collision occured

= h - h/3

= 2h /3

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