Question

A body is thrown from the top of a tower of height 114 meter with 20 meter per second .find the horizontal range of the body

Answer 1

A stone is dropped from the top of a tower that is 200 meters tall. At the same time, another stone is projected upward from the ground with the velocity of 50 meters/second. How do I find where and when both the stones meet?

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Alagappan Chidambaram, Mechanical Engineering Student

Answered Aug 6, 2017

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The stones meet after 4 seconds at a height of 121.6m from the bottom of the tower.

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Abhimanyu Kumar Bhagat, studies Bachelor of Technology in Electronics and Communications Engineering & IES/GATE Aspirants at Ilahi...

Answered Jul 14

Since,S=ut+1/2 at^2.

In case of free fall S=ut+1/2gt^2 eqn…1

Two cases arises here,

Case:1- stone is dropped from the tower,

given

S1=200 meters , u=0

So that, S1=ut+1/2 at^2 since it's case of free fall

So, S=ut+1/2gt^2 => S=1/2gt^2 eqn..2

Case:2 stone projected upward

S2=50t-1/2 gt^2. eqn ..3

Total distance S=S1+S2

Or, 200=S1+S2

Or, 200=1/2 gt^2+50t-1/2gt^2

=> 200=50t=> t=4sec.

Put t=4 in eqn 2.

We have, S1=0+1/2gt^2 => 1/2×9.81×4^2= 78.48m.

And S2=50×4–1/2×9.81×4×4

=200–78.48=121.52m

Both stones meet from ground in air at the distance of 121.52m in 4s.