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Projection Under uniform Acceleration

Question

A body is projected horizontally from the top of a hill with a velocity of 9.8 m/s. What time elapses before the vertical velocity is twice the horizontal velocity?

Moderate
Solution

Vy = 2Vx ; gt = 2u
t = \frac{{2u}}{g} = \frac{{2 \times 9.8}}{{9.8}} = 2s



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