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Grade 11IIT JEE Entrance Exam

from the top of tower of height 400m, a ball is dropped by the man,simultaneously from the base of the tower, another ball is thrown up with the velocity 50m/s.at what height they will meet from the base of the tower?(g=10m/s2)

Profile image of Nakshi Shah
8 Years agoGrade 11
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1 Answer

Profile image of Pavan sai vardhineedi
8 Years ago
Very simple trick
T=h/u
T= 400/50=8 sec
Now subs for x or y 
x for freely falling body and y for vertical projected
x=gt2/2=10×8×8/2=320m
We know h=x+y
400=320+you
y=80m
The answer will be 320m or 80m
 

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