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
2. Homework Equations
Xf=xi +vt + 0.5t^2
Vf= vi + at
The Attempt at a Solution
The First Stone.
d1 = -0.5g*t^2 = -4.9*(1.5)^2 = -11 m. =
Distance traveled by 1st stone after 1,5s.
Vf= 0 -9.8*1.5 = -14.7m/s
V = -14.7 m/s = Velocity of 1st stone after 1.5 s.
d1 = Vo*t + 0.5g*t^2
-200 = -11 - 14.7 t -4.9t^2
-4.9t^2 -14.7t +189=0
Using Quadratic Formula.
Tf = 4.89 s. = Fall time or time for each stone to reach Gnd.
The 2nd Stone.
d = Vo*t + 0.5g*t^2 = -200 m.
Vo*4.89 -4.9*(4.89)^2 = -200
Why Am I getting a negative answer the upward velocity should be positive (the top of the cliff is taking as the X=0 position)