Question

A projectile goes farthest from the earth when the angle an projection is:

Moderate

Solution

In vertical direction,
$u_{i} = u sinθ, u_{f} = 0 \begin{array}{l} u_{f}^{2} - u_{i}^{2} = - 2 gs \\ 0 - u^{2} \sin^{2} θ = - 2 gs \\ s = \frac{u^{2} \sin^{2} θ}{2 g} \end{array}$
where s is the maximum height.
For s to be maximum, sin $θ$ should be maximum, i. e., $θ$ = 90°.

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