Answer:
1.6 secs
Explanation:
In a circus act, an acrobat upwards from the surface of a trampoline
At that same moment another acrobat perched 9.0m above him
A ball is released from rest
While still in motion the acrobat catches the ball
He left the ball with a trampoline of 5.6m/s
Since the ball is falling downwards from a distance then acceleration will be negative
a= -9.8
s= d
s= 1/2at^2
= 1/2 × (-9.8)t^2
= 0.5× (-9.8)t^2
d = -4.9t^2
Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows
9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2
9 - 4.9t^2= 5.6t + (-4.9)t^2
9 - 4.9t^2= 5.6t - 4.9t^2
9= 5.6t
t= 9/5.6
t= 1.6 secs
