Question

In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9.0 m above him, releases a ball from rest. While still in flight, the acrobat catches the ball just as it reaches him.

If he left the trampoline with a speed of 5.6 m/s, how long is he in the air before he catches the ball? (express your answer in second)

Answer 1

Answer:

1.6 s

Explanation:

Let after time t both ball and acrobat meet in the mid air .

Distance traveled by acrobat

h₁ = u t -.5 g t²

= 5.6 t - 4.9 t²

Distance traveled by ball in time t , when it is released with initial velocity zero

h₂ = .5 g t²

= 4.9 t²

Now h₁ + h₂ = 9 m

5.6t - 4.9 t² + 4.9 t² = 9

5.6 t = 9

t =  $\frac{9}{5.6}$

1.6 s