A ball is dropped from rest from a height h above the ground.Another ball is thrown vertically upwards from the ground at the instant the first ball is released.Determine the initial speed of the second ball if the two balls are to meet at a height h/2 above the ground.
A)h/g
D)g/h

Question

A ball is dropped from rest from a height h above the ground.Another ball is thrown vertically upwards from the ground at the instant the first ball is released.Determine the initial speed of the second ball if the two balls are to meet at a height h/2 above the ground.
A)h/g 
D)g/h

Quizplus · Accepted Answer

Using conservation of energy, we can equate the potential energy lost by the first ball to the potential energy gained by the second ball at the meeting point. 
mgh = (1/2)mv^2 + (1/2)mg(h/2) 
h cancels out 
gh = (1/2)v^2 + (1/4)gh 
Simplifying, 
v^2 = 3gh/2 
v = sqrt(3gh/2)

Substituting h/g for t, and applying the Equation v = gt as the initial velocity of the second ball is zero, 
v = gt = sqrt(3gh/2) 
v = sqrt(3g(h/g)/2) 
v = sqrt(3g/2) * sqrt(h/g) 
v = 11ea821f_c6d2_0133_bbd8_af9f689a3041_TB1516_11_TB1516_11_TB1516_11

A Ball Is Dropped from Rest from a Height H