Question

A grasshopper can jump a maximum horizontal distance of 0.2m. If it spends negligible time on the ground, with what speed can it travel along the ground?
A. $1{\text{ m/s}}$
B. $2{\text{ m/s}}$
C. $3{\text{ m/s}}$
D. $4{\text{ m/s}}$

Answer 1

Projectile motion is a form of motion experienced by an object that is projected near the Earth’s surface and moves along a curved path under the action of gravity. The projectile has a single force that acts upon it, which is the force of gravity.
In this question, the maximum range of the grasshopper is given, so by using the maximum range formula, first find the speed of the grasshopper and then find the x-component of the obtained speed to find the traveling speed of the grasshopper along the ground.

Complete step by step answer:
The maximum horizontal distance of$R = 0.2m$

We know the range of a projectile is given by the formula
$R = \dfrac{{{u^2}\sin 2\theta }}{g} - - (i)$
In this formula, we can see the range is directly dependent on the speed and the angle of projection and to obtain maximum range the speed and the angle of projection must be maximum,
In equation (i)
$R \propto \sin 2\theta$
For maximum range $\sin 2\theta$ must be maximum and $\operatorname{Sin} 2\theta$ is maximum when $\theta = {45^ \circ }$ , since

$$\sin {90^ \circ } = 1 \\\ \sin 2\left( {{{45}^ \circ }} \right) = 1 \\\$$

$\therefore \theta = {45^ \circ }$

Now substitute the values $R = 0.2m$ and $\theta = {45^ \circ }$ in equation (i), hence we get

0.2 = \dfrac{{{u^2}\sin 2\left( {{{45}^ \circ }} \right)}}{{10}} \\\ 2 = {u^2} \times 1 \\\ u = \sqrt 2 \dfrac{m}{{{s^2}}} \\\ $$[Since$$g = 10{\text{ m/}}{{\text{s}}^2}$$] Since we are asked to find speed along the ground, hence the horizontal component of the speed will be

u\cos \theta = \sqrt 2 \times \cos \left( {{{45}^ \circ }} \right) \\
= \sqrt 2 \times \dfrac{1}{{\sqrt 2 }} \\
= 1\dfrac{m}{{{s^2}}} \\

Therefore, the speed at which grasshopper can travel along the ground Option (A) is correct **Note:** Students must note that sin function reaches to the maximum value 1 when the input angle is $${90^ \circ }$$ so for $$2\theta $$ to be maximizing the value of the angle $$\theta = {45^ \circ }$$