A stone is thrown horizontally with velocity √(2gh) from a top of a tower of hight 'h'. Find where it will strike the level ground through the foot of the tower. What will be it's striking velocity?

1 Answer
A08
Jun 2, 2018

Horizontal and vertical velocities are orthogonal to each other and therefore can be treated separately. We ignore air resistance.

  1. Vertical motion is free fall under gravity and is governed by kinematic expression
    #h=ut+1/2g t^2#
    Time #t# to reach ground is
    #h=0xxt+1/2g t^2#
    #=>t=sqrt((2h)/g)# ......(1)
    Final velocity as stone hits ground can be found with the help of kinematic expression
    #v=u+at#
    #=>v_(vf)=0+gxxsqrt((2gh))#
    #=>v_(vf)=sqrt(2gh)#

  2. Horizontal motion.
    Distance #R# moved during this time in the horizontal direction
    #R=sqrt (2gh)xxsqrt((2h)/g)=2h#

We see that modulus of final vertical velocity is equal to horizontal velocity. As such resultant velocity as stone hits ground will make an angle of #45^@# with the #y#-axis.
Also #v_R=sqrt((sqrt(2gh))^2+(sqrt(2gh))^2)#
#=>v_R=2sqrt(gh)#

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