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A ball is of all is allowed to fall from the top of a tower of 100m high At the same instant another ball is thrown vertically upward from the bottom of the tower with a velocity of 20 m per second when and where the two balls meet

Posted by Gourav Paliwal 8 years, 1 month ago

CBSE > Class 11 > Physics

1 answers

Ashish Soni 8 years, 1 month ago

Given height 100 m and velocity of 2nd ball 20 m per sec Now S=ut+1/2 at^2 Consider that they will meet at h point and h distance So H= ut+1/2 at^2 Solve it for 2nd ball and a= -9.8 m per second square For 1st ball 100-h =ut+1/2 at^2 Solve it take acceleration 9.8 m per sec sq.and u= 0 Now from both eq we get t=5 Then put t in eq 2 answer will come approx 22.5 m

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