First slide

Projectile motion

Question

In figure, find the horizontal velocity u (in ms^-1) of a projectile so that it hits the inclined plane perpendicularly. Given H = 6.25 m. $Take g = 10 {ms}^{- 2}$

Moderate

Solution

$v_{x} = u_{x} + a_{x} t$
$\Rightarrow 0 = ucos 30^{\circ} - gsin 30^{\circ} t$
$\Rightarrow t = \frac{u \sqrt{3}}{g}$
$S_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}$
$\begin{aligned} \Rightarrow - Hcos 30^{\circ} = - usin 30^{\circ} t \\ - \frac{1}{2} gcos 30^{\circ} t^{2} \end{aligned}$
$\Rightarrow - H \frac{\sqrt{3}}{2} = \frac{- u}{2} \frac{u \sqrt{3}}{g} - \frac{1}{2} g \frac{\sqrt{3}}{2} \frac{u^{2} 3}{g^{2}}$
$\Rightarrow u = \sqrt{2 gH / 5} = \sqrt{2 \times 10 \times 6 .25 / 5} = 5 {ms}^{- 1}$

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