Three balls are launched from the top of a building of height H. The balls m1 an

Question

Three balls are launched from the top of a building of height H. The balls m1 and m2 are launched with the same initial speed Vo with m1 making an angle of 45 degrees above the horizontal and m2 launched horizontally (i.e. 0 degrees to the horizontal); m3 is dropped straight vertically down from rest.

If m3>m2>m1 what will be the correct relationship between their final speeds when they hit the ground level (base level of the building)? Assume air resistance is negligible

Correct answer (V1=V2)>V3... confused as to why

Explanation / Answer

Here ,

as the initial speed is same for all the balls

kinetic energy is same for all the balls

for the speed at the bottom

as Vf^2 = Vo^2 + 2 * g * H

as Vo and h is same for all the balls

speed at the bottom is independent of mases of the ball

the relation between the speeds is

Vf1 = vf2 = vf3

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