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mylen
2 months ago
12

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.Required: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?
Physics
1 answer:
Keith_Richards [3.2K]2 months ago
6 0

Answer:

1.6 secs

Explanation:

During a circus performance, an acrobat jumps upward from a trampoline’s surface

At that precise moment, another acrobat at 9.0m above him releases a ball from rest

While still airborne, the acrobat catches the falling ball

He departed the trampoline with a speed of 5.6m/s

As the ball descends from a height, the acceleration is 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

Thus, the duration the acrobat remains in the air prior to 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

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