91Ó°ÊÓ

The takeoff distance for each jump attained by the kangaroo is $10\text{m}$and the angle of projection is $20°$.

When the kangaroo is at the maximum height from the ground, its final speed is zero and the time taken by the kangaroo is half of the total time before it again touches the ground.

The formula to calculate the horizontal distance travelled $\left(S\right)$by the kangaroo is given by

$S=v_0\cos \mathrm{θ}t.....................\left(1\right)$

Here, $v_0,\mathrm{θ},t$are the takeoff speed, the angle of projection and the total time taken by the kangaroo.

Substitute $10$for $S$and $20°$for $\mathrm{θ}$into equation (1) and solve to obtain the time taken by the kangaroo.

role="math" localid="1647409257488" $\begin{array}{rcl}10& =& v_0\cos 20°t\\ t& =& \frac{10}{v_0\cos 20°}\\ & ≈& \frac{10}{0.94v_0}..............\left(2\right)\end{array}$

The formula to calculate the final speed of the kangaroo when it attains the maximum height from the ground is given by

$v=v_0\sin \mathrm{θ}-\frac{gt}{2}............\left(3\right)$

Substitute $0$for $v$and the value of the time from equation (2) into equation (3) and solve to obtain the expression for the takeoff speed.

role="math" localid="1647409542034" $\begin{array}{rcl}0& =& v_0\sin \mathrm{θ}-\frac{g}{2}\left(\frac{10}{0.94v_0}\right)\\ v_0& =& \sqrt{\frac{10g}{2×0.94×\sin \mathrm{θ}}}...............\left(4\right)\end{array}$

Substitute $9.80{\text{m/s}}^2$for $g$and $20°$for $\mathrm{θ}$into equation (4) to calculate the required speed.

$\begin{array}{rcl}{v}_0& =& \sqrt{\frac{10×9.80{\text{m/s}}^2}{2×0.94×\sin 20°}}\\ & ≈& 12.38\text{m/s}\end{array}$

Substitute $12.38\text{m/s}$for $v_0$into equation (2) to calculate the time taken by the kangaroo.

$\begin{array}{rcl}t& =& \frac{10\text{m}}{0.94×12.38\mathrm{m}/\mathrm{s}}\\ & ≈& 0.86\text{s}\end{array}$

The formula to calculate the maximum height attained by the kangaroo is given by

role="math" localid="1647409922073" $h=v_0\sin \mathrm{θ}-\frac{1}{2}g{\left(\frac{t}{2}\right)}^2.............\left(5\right)$

Substitute the proper values of the parameters into equation (5) to calculate the required maximum height.

$\begin{array}{rcl}h& =& 12.38\text{m/s}×\sin 20°-\frac{1}{2}×9.80{\text{m/s}}^2×{\left(\frac{0.86\text{s}}{2}\right)}^2\\ & ≈& 0.90\text{m}\end{array}$