Projection Under uniform Acceleration

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Question

Two bodies are thrown from the same point with the same velocity of 50 ms^–1. If their angles of projection are complimentary angles and the difference of maximum heights is 30 m, their maximum heights (g = 10 ms^-2)

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Solution

$\large {H_1} - {H_2} = 30;\;{H_1} + {H_2} = \frac{{{u^2}}}{{2g}} = \frac{{50 \times 50}}{{2 \times 10}} = 125$
$\large \therefore 2{H_1} = 155 \Rightarrow {H_1} = 77.5m;\;{H_2} = 47.5m$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

Two bodies are thrown from the same point with the same velocity of 50 ms–1. If their angles of projection are complimentary angles and the difference of maximum heights is 30 m, their maximum heights (g = 10 ms-2)

Ready to Test Your Skills?

free study material

Two bodies are thrown from the same point with the same velocity of 50 ms^–1. If their angles of projection are complimentary angles and the difference of maximum heights is 30 m, their maximum heights (g = 10 ms^-2)