# Question #3c4aa

Nov 7, 2016

Stone is projected with a velocity $u = 20 \text{ m/s}$ at angle of projection $\alpha = {48}^{\circ}$ with the horizontal from the edge of a cliff at the height 60.0m above the surface of the sea.

The vertical component of velocity of projection is $u \sin \alpha$ and horizintal component is $u \cos \alpha$

If the vertical component of velocity be v after covering the donward vertical displacement $h = - 60 m$ then we can write

${v}^{2} = {\left(u \sin \alpha\right)}^{2} + 2 \times g \times h$

The horizontal component of velocity remains unchanged.It is $= u \cos \alpha$

So the resultant velocity $\left({v}_{r}\right)$ of the stone at sea level is

${v}_{r} = \sqrt{{v}^{2} + {\left(u \cos \alpha\right)}^{2}}$

$= \sqrt{{\left(u \sin \alpha\right)}^{2} + 2 g h + {\left(u \cos \alpha\right)}^{2}}$

$= \sqrt{{u}^{2} + 2 g h}$

$= \sqrt{{20}^{2} + 2 \left(- 9.8\right) \left(- 60\right)}$

$= 39.7 \text{ m/s}$

The velocity with which the stone hits the sea is
$= 39.7 \text{ m/s}$