The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of$\mathbf{58}\mathbf{.}\mathbf{0}\mathbf{°}$above the horizontal, some of the tiny critters have reached a maximum height of 58.7 cm above the level ground. (See Nature, Vol. 424, July 31, 2003, p. 509.) (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?

The given data can be listed below.

The angle above the ground level is $58.0°$.

The maximum height of the critter can reach is 58.7 cm .

The speed is a measurement of how quickly something changes location in relation to time.

The value of takeoff speed is given by,

${\mathrm{v}}_{\mathrm{y}}^2={\mathrm{v}}_{0\mathrm{y}}^2-2\mathrm{g}\left(\mathrm{y}-{\mathrm{y}}_0\right)$

Here, $v_{0y}$ is the initial velocity,g is the acceleration due to gravity and $\left(y-y_0\right)$is the distance traveled by the leap.

Substitute all the values in the above expression.

localid="1667623358175" $$\begin{gathered} -{\mathrm{v}}_{0\mathrm{y}}^2{\sin }^{2}58°=-{\mathrm{v}}_{\mathrm{y}}^2-2\mathrm{g}\left(\mathrm{y}-{\mathrm{y}}_0\right) \\ {\mathrm{v}}_{0\mathrm{y}}=\sqrt{\frac{2\mathrm{g}\left(\mathrm{y}-{\mathrm{y}}_0\right)+{\mathrm{v}}_{\mathrm{y}}^2}{{\sin }^{2}58°}} \\ =4\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the takeoff speed of the froghopper is 4 m/s .

The distance covered by froghopper is given by,

$\mathrm{y}={\mathrm{y}}_0+{\mathrm{v}}_{0\mathrm{y}}\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2$

Here, ${\mathrm{y}}_0$is the vertical distance of froghopper, $v_{0y}$is the vertical component of velocity, t is the time taken.

Substitute all the values in the

$$\begin{gathered} 0=0+{\mathrm{v}}_0\mathrm{sin\alpha t}-\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{t}=\frac{2{\mathrm{v}}_0\mathrm{sin\alpha t}}{\mathrm{g}} \end{gathered}$$

The horizontal range of the froghopper is given by,

role="math" localid="1665044557236" $\mathrm{R}=\left(v_0\cos \alpha \right)t$

Substitute the $\frac{2{\mathrm{v}}_0\mathrm{sin\alpha t}}{\mathrm{g}}$for t in the above expression.

$\mathrm{R}=\left({\mathrm{v}}_0\mathrm{cos\alpha }\right)\left(\frac{2{\mathrm{v}}_0\sin \mathrm{\alpha t}}{\mathrm{g}}\right)$

Substitute all values in the expression for range.

$$\begin{gathered} \mathrm{R}=\left(4\mathrm{m}/\mathrm{s}\right)\left(\frac{2\times 4\mathrm{m}/\mathrm{s}\left(\sin 58°\right)}{9.8\mathrm{m}/{\mathrm{s}}^2}\right) \\ =1.4\mathrm{m} \end{gathered}$$

Thus, the horizontal distance the froghopper cover for world-record leap is 1.4 m .