Posted by chibuzor on Wednesday, January 30, 2013 at 6:00am.
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

mechanics  Henry, Friday, February 1, 2013 at 5:27pm
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.

mechanics  Henry, Friday, February 1, 2013 at 5:43pm
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.
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