0
505views
A projectile is thrown at an angle and another is thrown at (90- ) from the same point both with the velocities 78.4ms-1. The second reaches 36.4m higher than the first. Find the individual heights.
0
16views

Solution:

Given,

Angle of projection of first projectile, $=\theta$

Angle of projection of second projectile, $=(90-\theta)$

Velocity of projection, $u=78.4 \mathrm{~ms}^{-1} \\$

and, $\mathrm{H}_{2}=\mathrm{H}_{1}+36.4 \mathrm{~m} \\$

$\mathrm{H}_{2}-\mathrm{H}_{1}=36.4 \mathrm{~m} \\$

Where, $\mathrm{H}_{1}$ is the maximum height of first one and $\mathrm{H}_{2}$ is the maximum height of second one.

$\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}} \\$ and , $\mathrm{H}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}(90-\theta)}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{~g}} \\$

Adding, $$\mathrm{H}_{2}+\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{~g}}+\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\\$$

\begin{aligned} &=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}}=\frac{78.4 \times 78.4}{2 \times 9.8}=313.6 \\\\ 2 \mathrm{H}_{2} &=36.4+313.6=350 \\\\ \mathrm{H}_{2} &=175 \mathrm{~m} \\\\ \end{aligned}

Also, $$\mathrm{H}_{1}=\mathrm{H}_{2}-36.4=175-36.4 \\$$

$$\mathrm{H}=138.6 \mathrm{~m}\\$$