Question

Two bodies are thrown from the same point with the same velocity of 50 m/s. If their angles of projection are complimentary to each other and the difference in maximum heights is 30 m, are the maximum heights?

Answer 1

Step 1: Given data

Velocity $u=50\frac{\text{m}}{\text{s}}$.

${\theta }_1+{\theta }_2={90}^{°}$

$H_{max1}-H_{max2}=30\text{m}$

Step 2: To Find

Maximum heights.

Step 3: Calculate the angle of projection

Maximum height formula

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

Here,

$u$ is velocity, $\theta$ is the angle of projection and $g$ is the gravitational constant.

$$\begin{gathered} H_{max1}-H_{max2}=30 \\ \frac{u^2{\sin }^2\left({\theta }_1\right)}{2g}-\frac{u^2{\sin }^2\left({90}^{°}-{\theta }_1\right)}{2g}=30 \\ \frac{{50}^2}{2\times 10}\left({\sin }^2{\theta }_1-{\cos }^2{\theta }_1\right)=30 \\ \cos 2{\theta }_1=\frac{600}{{50}^2} \\ {\theta }_1={38}^{°} \end{gathered}$$

Step 4: Calculate heights

$$\begin{gathered} H_{max1}=\frac{{\left(50\right)}^2{\sin }^2{38}^{°}}{2\times 10} \\ =47.5\text{m} \end{gathered}$$

$$\begin{gathered} H_{max2}=47.5\text{m}-30\text{m} \\ =17.5\text{m} \end{gathered}$$

Hence, the heights are $47.5\text{m},17.5\text{m}$.