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In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

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The horizontal and vertical components of the velocity decrease due to air resistance. So, as a result, the max height also becomes lesser than in the ideal case.

Now, R=u^{2}\sin 2\; \theta/g and max height H=u^{2}\sin^{2} \; \theta/2g

So, h1<h2\;and\; R1<R2

But, in the second case when h1<h2, due to smaller time taken to rise, the overall time of flight for both cases stands equal

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