mechanics
posted by chibuzor on .
a stone is dropped from a cliff 200m high.if a second stone is thrown verticall upwards 1.50seconds after the first was released.strike the cliff base at the same instant as the first stone.with what velocity was the second stone thrown

The First Stone.
d1 = 0.5g*t^2 = 4.9*(1.5)^2 = 11 m. =
Distance traveled by 1st stone after
1,5 s.
V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*11 = 215.6
V = 14.68 m/s = Velocity of 1st stone after 1.5 s.
d1 = Vo*t + 0.5g*t^2 = 20011.
16.68t + 4.9t^2 = 189
4.9t^2 + 16.68t  189 = 0.
Use Quadratic Formula.
Tf = 4.74 s. = Fall time or time for each stone to reach Gnd.
The 2nd Stone.
d = Vo*t + 0.5g*t^2 = 200 m.
Vo*4.74 + 4.9*(4.74)^2 = 200
4.74Vo + 110.1 = 200
4.74Vo = 200110.1 = 89.9
Vo = 19 m/s. 
Correction(TYPO).
Change 16.68 to 14.68. Then Tf will be
4.89 s.
Change all 4.74s to 4.89.
Vo = 89.9/4.89 = 18.4 m/s. = Initial velocity of 2nd stone.