Question

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

Moderate

Solution

Initial velocity is $(6 \hat{i} + 8 \hat{j}) m / s (given) .$
Magnitude of velocity of projection is
$u = \sqrt{u_{x}^{2} + u_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 m / s$
Angle of projection
$\begin{array}{l} \tan θ = \frac{u_{y}}{u_{x}} = \frac{8}{6} \\ \Rightarrow \sin θ = \frac{4}{5} and \cos θ = \frac{3}{5} \end{array}$
Now the horizontal range is,
$\begin{array}{l} R = \frac{u^{2} \sin 2 θ}{g} = \frac{u^{2} 2 \sin θcos θ}{g} \\ = \frac{(10)^{2} \times 2 \times (4 / 5) \times (3 / 5)}{10} = 9 .6 m \end{array}$

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