Question #3c4aa

1 Answer
Nov 7, 2016

Stone is projected with a velocity #u=20" m/s"# at angle of projection #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 #usinalpha# and horizintal component is #ucosalpha#

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

#v^2=(usinalpha)^2+2xxgxxh#

The horizontal component of velocity remains unchanged.It is #=ucosalpha#

So the resultant velocity #(v_r)# of the stone at sea level is

#v_r=sqrt(v^2+(ucosalpha)^2)#

#=sqrt((usinalpha)^2+2gh+(ucosalpha)^2)#

#=sqrt(u^2+2gh)#

#=sqrt(20^2+2(-9.8)(-60))#

#=39.7" m/s"#

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