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
Horizontal and vertical velocities are orthogonal to each other and therefore can be treated separately. We ignore air resistance.
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Vertical motion is free fall under gravity and is governed by kinematic expression
h=ut+1/2g t^2h=ut+12gt2
Timett to reach ground is
h=0xxt+1/2g t^2h=0×t+12gt2
=>t=sqrt((2h)/g)⇒t=√2hg ......(1)
Final velocity as stone hits ground can be found with the help of kinematic expression
v=u+atv=u+at
=>v_(vf)=0+gxxsqrt((2gh))⇒vvf=0+g×√(2gh)
=>v_(vf)=sqrt(2gh)⇒vvf=√2gh -
Horizontal motion.
DistanceRR moved during this time in the horizontal direction
R=sqrt (2gh)xxsqrt((2h)/g)=2hR=√2gh×√2hg=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
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