# Falling Objects| Kinematics| Jumping Kangaroo|College Physics

### A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Solution:

PART A

We are going to compute for the initial velocity given that $y=2.50\:m$; we also know that the velocity at the top is 0. So, we can use the formula $v^2=\left(v_0\right)^2+2ay$. Solve for $v_0$.

$v_0=\sqrt{v^2-2ay}$

Substitute the given values:

$v_0=\sqrt{\left(0\right)^2-2\left(-9.80\:m/s^2\right)\left(2.50\:m\right)}$

$v_0=7.00\:m/s$

Therefore, the initial velocity of the kangaroo is 7.00 m/s.

PART B

Next, we solve for the time using the formula

$v=v_0-gt$

Solve for the time it takes to reach the maximum height of 2.50 m.

$t=\frac{v_0-v}{g}$

Substitute the given values

$t=\frac{7.00\:m/s-0}{9.80\:m/s^2}=0.714\:s$

We know that the time it takes to reach the highest peak is equal to the time it takes to go back to the original position. The total time in the air is

$t_{total}=2\left(0.714\:s\right)=1.428\:s$

Therefore, the kangaroo is in the air for 1.428 seconds.