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.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?

Answer 1

The acrobat is in the air for the same amount of time as it takes for the ball to move 9.0 m downward. Using kinematic equations for both the acrobat's upward motion and the ball's downward motion, we can solve for the time they both are in the air, which is when they meet.

Step-by-step explanation:

To calculate how long the acrobat is in the air before he catches the ball, we need to consider the motion of both the acrobat and the ball. Let's use the kinematic equations of motion and the given values. The acrobat rebounds upwards with an initial speed of 5.6 m/s. Using the kinematic equation final velocity squared (v^2) = initial velocity squared (u^2) + 2 * acceleration (a) * displacement (s), where a is the acceleration due to gravity (which is -9.8 m/s² since it's directed downwards), we can solve it for time (t).

For the ball, released from rest above, its displacement (s) after time t is given by s = ut + 1/2 * a * t^2, where u is the initial velocity (0 m/s), and a is again the acceleration due to gravity (-9.8 m/s²). Given that the ball is at a height of 9.0 m, we can use this equation to find the time t it takes for the ball to reach the acrobat.

Since both the acrobat and ball are subject to the same acceleration and must meet at the same point, they will have the same t value when they meet. Calculating for t using the above equations will give us the time the acrobat is in the air before he catches the ball. Note that this is a simplification assuming constant gravitational acceleration and neglecting air resistance and other possible forces.

Answer 2

1.6 secs

Step-by-step 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