Question

A Kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to the ground?

Short answer

The kangaroo was in the air for 1.09 s before it returned to the ground.

Step-by-step solution

Step 1. Velocity at the highest point

At an instance of time, when an object is at the maximum height, the vertical component of its velocity must be zero, as it is not moving in the upward direction anymore.

Step 2. Given data and assumption

The maximum height achieved by the kangaroo, $h=1.45\mathrm{m}$.

Assume that the kangaroo stays in the air for $t_{\circ }$time, its initial velocity is u, and the acceleration due to gravity is localid="1654716117747" $\mathrm{g}=-9.81 {\mathrm{ms}}^{-2}$. (Taking the upward direction as positive)

Step 3. Calculating the initial velocity of the kangaroo

When the kangaroo reaches its maximum height, its velocity will become zero. Thus, using the third equation of motion,

$0^2-u^2=2\mathrm{g}h$.

Substituting the values in the above equation,

localid="1654716139986" $0-u^2=2\times \left(-9.8\right)\times 1.45 \frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}$.

Solving for the value of u,

localid="1654716162369" role="math" $u=5.33 {\mathrm{ms}}^{-1}$.

Step 4. Calculating the time in the air

From the second equation of motion, you can write the displacement of the kangaroo at time t as

$s=u\times t+\frac{1}{2}\mathrm{g}{t}^2$.

Here, as the initial and final positions are the same, the displacement of the kangaroo is zero. Thus,

localid="1654716260998" $0=5.33 \frac{\mathrm{m}}{\mathrm{s}}\times t_{\circ }+\frac{1}{2}\times \left(-9.81 \frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\times t_{\circ }^2$.

Solving the above quadratic equation and neglecting root t = 0,

localid="1654716268592" $t_{\circ }=1.09 \mathrm{s}$.

Thus, the total time for which the kangaroo was in the air is 1.09 s.