Question

A projectile is projected with initial velocity $(6 \hat{i} + 8 \hat{j}) m / \sec$ .If $g = 10 {ms}^{- 2}$ , then horizontal range is

Easy

Solution

Here, $ucos θ = 6, usin θ = 8 and g = 10$
$\begin{array}{l} R = \frac{u^{2} \sin 2 θ}{g} = \frac{2 u^{2} \sin θcos θ}{g} \\ = \frac{2 (usin θ) (ucos θ)}{g} = \frac{2 \times 8 \times 6}{10} = 9 .6 m \end{array}$

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