First slide

Projection Under uniform Acceleration

Question

A body projected at 45° with a velocity of 20 m/s has 10 m decrease in range due to air resistance. Then the final range is (g = 10ms^-2)

Moderate

Solution

In the absence of air resistance $\large R = \frac{{{u^2}\sin 2\alpha }}{g} = \frac{{{{(20)}^2}\sin (2 \times {{45}^0})}}{{10}} = 40m$
$\large \therefore$ ΔR = 40m - 10m = 30m

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