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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.
written 3.6 years ago by gravatar for prajapatijaimin prajapatijaimin • 3.3k •   modified 3.6 years ago
engineering physics
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written 3.6 years ago by gravatar for prajapatijaimin prajapatijaimin • 3.3k

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 …

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