A grasshopper hops along a level road. On each hop, the grasshopper launches itself at an angle${\theta }_0={45}^{\mathrm{o}}$and achieves a range$R=0.80\mathrm{m}$. What is the average horizontal speed of the grasshopper as it hops along the road? Assume that the time spent on the ground between the hop is negligible.

The horizontal speed of a projectile remains constant as the projectile experiences no acceleration in the horizontal direction.

It is defined as the ratio of the total horizontal displacement to the time taken.

The range of the grasshopper is $R=0.80\mathrm{m}$.

The launch angle of the grasshopper is $\theta ={45}^{\mathrm{o}}$.

The acceleration due to gravity is $g=9.8\mathrm{m}/{\mathrm{s}}^2$.

Let the initial launch speed of the grasshopper be u.

The formula for the horizontal range is given by:

$R=\frac{u^2\sin 2\theta }{g}\dots \left(i\right)$

The initial speed from equation (i) is given as:

$\begin{array}{c}u=\sqrt{\frac{gR}{\sin 2\theta }}\\ =\sqrt{\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.80\mathrm{m}\right)}{\sin {90}^{\mathrm{o}}}}\\ =2.8\mathrm{m}/\mathrm{s}\end{array}$

The time between the jumps of the grasshopper is negligible. Therefore, its initial speed is the horizontal component of the launch speed.

The speed along the x-direction is given by:

$\begin{array}{c}{u}_{\mathrm{x}}=u\cos \theta \\ =\left(2.8\mathrm{m}/\mathrm{s}\right)\left(\cos {45}^{\mathrm{o}}\right)\\ =\left(2.8\mathrm{m}/\mathrm{s}\right)\left(\frac{1}{\sqrt{2}}\right)\\ \approx 2.0\mathrm{m}/\mathrm{s}\end{array}$