Answer to Question #263511 in Mechanical Engineering for H..

Question

Stone A is hurled vertically upwards from a tower of height h with a speed vA0 . Simultaneously, stone B is thrown vertically upwards from the ground with a speed of vB0.

Given:

h=30m ; vA0=15m/s ; vB0=25m/s ; g=9.81m/s2 Find:

1. At which height and time do the stones fly past another?

2. At this time, what is the speed of stone A relative to stone B?

Answer 1

Given Initial velocity of ball, u=25 m/sLet the maximum height reached and time taken to reach that height be H and t respectively.

1-

Velocity of the ball at maximum height is zero, v=0

$v ^2 −u ^2 =2aH$

$0−(49) ^2 =2×(−9.8)×H$

H=122.5 m   

2-

v=u+at

0=49−9.8t

t=5s

Total time taken by ball to return to the surface, T=2t=10 s