Question

Two stones are projected with the same speed but making different angles with the Horizontal. Their horizontal ranges are equal. The angle of projection of the stones are$\frac{\mathrm{\pi }}{3}$, and$\frac{\mathrm{\pi }}{6}$. The maximum height reached by the first stone is $102\mathrm{meters}$. Calculate the maximum height reached by the other in meters.

Answer 1

Step 1: Given information

The angle of projection

$$\begin{gathered} {\theta }_1=\frac{\mathrm{\pi }}{3} \\ {\theta }_2=\frac{\mathrm{\pi }}{6} \end{gathered}$$

Here, ${\theta }_1$ is the angle of projection for the stone 1, and ${\theta }_2$ is the angle of projection for the stone 2.

Step 2: Formula used

1. The equation for the maximum height reached by the projectile can be written as follows:

$H=\frac{u^2{\sin }^2\theta }{2g}$

Here, $H$ is the maximum height reached, $u$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.

2. Since, the initial velocity is the same for the two stones, and acceleration due to gravity is a constant, the maximum height depends on the angle of projection only.

Hence, the ratio of maximum height reached can be expressed as follows:

$\frac{{\mathrm{H}}_1}{{\mathrm{H}}_2}=\frac{{\sin }^2{\mathrm{\theta }}_1}{{\sin }^2{\mathrm{\theta }}_2}$

Here,

${\mathrm{H}}_1$ and ${\mathrm{H}}_2$ are the maximum height reached by the stone $1$ and stone $2$ respectively.

${\mathrm{\theta }}_1$, ${\mathrm{\theta }}_2$ are the angle of projections for stone 1 and stone 2 respectively.

Step 3: Calculation of maximum height reached by the second stone.

$$\begin{gathered} \frac{102\mathrm{m}}{H_2}=\frac{{\sin }^2\left({\displaystyle \frac{\pi }{3}}\right)}{{\sin }^2\left({\displaystyle \frac{\pi }{6}}\right)} \\ \frac{102\mathrm{m}}{H_2}=\frac{{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}{{\left({\displaystyle \frac{1}{2}}\right)}^2} \\ {H}_2=\frac{102\mathrm{m}}{3} \\ {H}_2=34\mathrm{m} \end{gathered}$$

Hence, the maximum height reached by the second stone is $34\mathrm{m}$.