Two ball are thrown simultaneously, A vertically upward with a speed of 20m/s from the ground , and B vertically downward from a height of 40m with the same speed along the same line of motion . At what point do two balls meet
Let h be the distance from the ground where the collision takes place.
Distance from top of tower up to the collision point =40-h
Let collision took place after time t then:
For ball A
"h=20t-\\frac{1}{2}gt^2\u2026\u2026(1)"
For ball B
"40-h=20t+\\frac{1}{2}gt^2\u2026\u2026(2)"
Adding (1) and (2) we get
40=40t
Therefore t=1 sec
Putting value of t in (1) we get
"h=(20\u00d71)-(\\frac{1}{2}\u00d79.8\u00d71^2)=15.1m"
"( taking :g= 9.8 m \/s^2)"
h=15.1 metres
Therefore the two ball meet at 15.1m above the ground
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