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Grade 11Mechanics

A man is standing on top of a building 100m high. He throws two balls vertically, one at t=0 and other after a time interval (less than 2s).The later ball is thrown at a velocity half of the first. The vertical gap between first and second ball is 15 m at t = 2s. The gap is found to remain constant. The velocities at which the balls were thrown are?

Profile image of Piyush Upadhyay
8 Years agoGrade 11
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1 Answer

Profile image of Arun
8 Years ago
Dear Piyush
 
The only way for the gap to remain constant is if they have the same velocity for all times after that 2 seconds. In fact, this must be true from the moment the second ball was thrown, otherwise their velocities would never be equal.

So if their velocities are equal at all times, that means when the man threw the second ball, he had to have given it an initial velocity equal to the first balls velocity at that time. Also, since the distance is always 15m he threw the second ball when the first was 15m away. And I almost forgot the second is thrown half as fast as the first. 

So from that I can tell you that in 15m the first balls velocity went down by half 
Vf^2 = Vi^2 + 2*a*d 
(.5Vi)^2 = Vi^2 - 2*9.81*15 
294.3 = .75Vi^2 
Vi = 19.81 m/s 

which means the second one was thrown at 
Vi2 = .5*Vi1 
Vi2 = 9.90 m