A body freely falling from a height 50m and simultaneously another body is thrown
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Let t be the time when two bodies will meet.
As the body meets at a height of 10 m from the earth surface.
Then, the falling body will move distance of 40 m.The body thrown upward will move the distance of 10 m.
Taking downward motion of the body.
We have,
Initial velocity, u=0
Acceleration, a=g=9.8 m s^−2
S=ut+1/2at^2
⇒40=0×t+1/2×9.8×t^2
⇒40=4.9t^2 ........(1)
⇒t^2=40/4.9=8.16
⇒t=2.86 s
Taking vertical upward motion of the ball.
We have,
S=u×2.86+1/2×(−9.8)×t^2
⇒10=2.86u−4.9t^2
⇒10=2.86u−40
⇒2.86u=50
⇒u=50/2.86=17.48 ms^−1
As the body meets at a height of 10 m from the earth surface.
Then, the falling body will move distance of 40 m.The body thrown upward will move the distance of 10 m.
Taking downward motion of the body.
We have,
Initial velocity, u=0
Acceleration, a=g=9.8 m s^−2
S=ut+1/2at^2
⇒40=0×t+1/2×9.8×t^2
⇒40=4.9t^2 ........(1)
⇒t^2=40/4.9=8.16
⇒t=2.86 s
Taking vertical upward motion of the ball.
We have,
S=u×2.86+1/2×(−9.8)×t^2
⇒10=2.86u−4.9t^2
⇒10=2.86u−40
⇒2.86u=50
⇒u=50/2.86=17.48 ms^−1
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