Find the average velocity of a projectile between the the instants it crosses half the maximum height.It is projected with a speed u at an angle theta with the horizontal.
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Total of flight T = (2u sinθ)/g Average velocity = (change in displacement)/time From the figure, it can be said AB is horizontal. So there is no effect of vertical component of the velocity during this displacement. So because the body moves at a constant speed of ‘u cos θ’ in horizontal direction. The average velocity during this displacement will be u ‘cos θ ‘ in the horizontal direction.
In y-direction, it has accelerated motion and in the x-direction, it has a constant speed u cosA where A is angle of projection.Average velocity in x-direction will be u cosA as average of any constant is constant itself. In y-direction net displacement will be zero and hence avg velocity zero.so net avg velocity is ucosA only.
In y-direction, it has accelerated motion and in the x-direction, it has a constant speed u cosA where A is angle of projection.Average velocity in x-direction will be u cosA as average of any constant is constant itself. In y-direction net displacement will be zero and hence avg velocity zero.so net avg velocity is ucosA only.
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