Question #3c4aa

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Question #3c4aa

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Answer 1 · 2016-11-07T05:32:29

Stone is projected with a velocity $u = 20 m/s$ $u=20" m/s"$ at angle of projection $α = 48^{\circ}$ $alpha=48^@$ 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 $u sin α$ $usinalpha$ and horizintal component is $u cos α$ $ucosalpha$

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

$v^{2} = {(u sin α)}^{2} + 2 \times g \times h$ $v^2=(usinalpha)^2+2xxgxxh$

The horizontal component of velocity remains unchanged.It is $= u cos α$ $=ucosalpha$

So the resultant velocity $(v_{r})$ $(v_r)$ of the stone at sea level is

$v_{r} = \sqrt{v^{2} + {(u cos α)}^{2}}$ $v_r=sqrt(v^2+(ucosalpha)^2)$

$= \sqrt{{(u sin α)}^{2} + 2 g h + {(u cos α)}^{2}}$ $=sqrt((usinalpha)^2+2gh+(ucosalpha)^2)$

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

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

$= 39.7 m/s$ $=39.7" m/s"$

The velocity with which the stone hits the sea is
$= 39.7 m/s$ $=39.7" m/s"$

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