Question

a man standing on the top of building 100vmetre high. he throws 2 balls one at t = 0 and other after an interval of less than 2 seconds. the later ball is thrown or a velocity of the half the first the vertical gap bw the first and second ball is 15 m at t = 2 seconds . the gap is found to remain constant and calculate the velocity with which the balls were thrown and the exact time interval between their throw

Answer 1

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