First slide
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Question

Assertion The maximum height of projectile is always 25% of the maximum range.
Reason For maximum range, projectile should be projected at 90°.                     

Moderate
Solution

To obtain maximum range, angle of projection must be 45°, i.e., θ = 45°.

Rmax=u2sin2×45°g=u2g

  Hmax=u2sin245°2g=u24g=Rmax4

So, Hmax is 25 % of Rmax

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