PHYSICS HELP NEEDED!!!!!!!!!!?

Question

A stone is dropped from the roof of a building; 2.00s after that, a second stone is thrown straight down with an initial speed of 25.0m/s, and the two stones land at the same time. a) How long did it take the first stone to reach the ground? b) How high is the building? c) What are the speeds of the two stones just before they hit the ground?

Answer 1

Distance traveled by 1st stone: (Take g to be 10)

S = ut + 0.5at^2

S1 = 0.5(10)(T^2) = 5T^2

Distance traveled by 2st stone:

S2 = 25t + 0.5(10)(t^2) = 25t + 5t^2

But T = t+2 and S1 = S2

5(t+2)^2 = 25t + 5t^2

t = 4s

S2 = 25(4) + 5(16) = 180m

S1 = 5(4+2)^2 = 180m (just to check)

For the First stone:

V = u + at = 0 + 10(6) = 60m/s

For 2nd stone

V = 25 + 10(4) = 65m/s

Answer 2

a) -0.5(9.8)*t^2 = -25(t-2)-0.5(9.8)(t-2)^2

-4.9t^2 = -25t+50-4.9(t^2-4t+4)

0= -25t+50+19.6t - 19.6

5.4t = 30.4

t= 5.62962963 s

b) h= -4.9(5.62962963)^2

   h= -155.2943759

the building is 155.2943759 m high

c) speed of first stone = at = 9.8*5.62962963 = 55.17037037 m/s

speed of second stone= v+at= 25+9.8*3.62962963 = 60.57037037 m/s

hope this helps