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 = 200-11.

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 = 200-110.1 = 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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