First slide

Projection Under uniform Acceleration

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)

Moderate

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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