here's a math question.?

Question

a stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 3.3 seconds later with an initial speed of 64.68 m/s. they hit the ground at the same time. the acceleration due to gravity is 9.8.
How long does it take the first stone to reach the ground?

Answer 1

3.3 second longer than the second stone

Answer 2

The fall distance is the same for both stones. Call that s. For the first stone, s = .5gt1^2, where t1 is the fall time for the first stone. For the second stone, s = v0*t2 + .5gt2^2, where v0 is the initial velocity and t2 the fall time for the second stone. Both eqs are equal to s, so they are equal to each other:

.5ft1^2 = v0*t2 + .5gt2^2.

The second time t2 starts 3.3 sec after t1, so if they hit the ground at the same time, t1 must be 3.3 seconds more than t2:

t1 = t2 + 3.3, or t2 = t1 - 3.3;

Substitute for t2 in the first eq:

.5gt1^2 = v0*(t1 - 3.3) + .5g*(t1 - 3.3)^2

Multiply out the terms

.5gt1^2 = v0*t1 - 3.3*v0 + .5gt1^2 - .5g*3.3*t1 + .5*g*3.3^2

The t1^2 terms cancel out, leaving only first order t1 terms:

.5*g*3.3*t1 - v0*t1 = -3.3*v0 + .5*g*3.3^2

t1 = (-3.3*v0 + .5*g*3.3^2)/(.5*3.3*g - v0)

t1 = 3.3 * (.5*g*3.3 - v0)/(.5*3.3*g - v0) = 3.3 sec

(EDITED)

Answer 3

holy ****. this is physics. man did i hate that class last year. but i'm glad it's over with and it's funny cause i look back now and think it's easy. anyways, i don't really feel like answering the actual question. sorry. but i hope you figure it out.