physics 2 part question?

Part 1

s = ut + (at^2)/2

s is same for both stones

but second stone takes 2.4 s less than first

u (initial speed of second stone) = 47.04

u first stone = 0

therefore:

47.04 t + (9.8 t^2) / 2 = (9.8(t + 2.4)^2) / 2

Simplifying:

27.44 t = 19.6

t = 0.714 s

Time for first stone = 2.4 + 0.714

= 3.11 s

Part 2:

Height = s = (9.8 (3.11)^2)/2

= 47.524 m

= 47.5 m

Hans' initial equation looks okay. (t = the time in seconds for the SECOND (thrown) stone to get to the ground.)

I think the problem is in the simplifying of the eqtn. (That usually is grungy work.)

I get 1.2 seconds for t. The first stone of course takes 3.6

seconds. The cliff height becomes 63.5 meters.

ok, i think of i will help you with this undertaking. (nonetheless it relatively is basically the 2nd question i've got responded on yahoo). right this is a physics formulation I basically love. Vf^2=Vi^2+2A*D. in actuality, the suitable velocity squared equals the preliminary velocity squared + 2 circumstances acceleration circumstances distance. i might propose changing a million.6 cm to .0016 meters first. fixing for A, we've: Acceleration= (velocity very final^2 minus velocity preliminary^2)/(2*.0016) ^2 potential to sq.,btw. i'm going to permit you paintings that out your self. For area 2, you ought to use the formulation D=(a million/2) A*T^2 seem on the internet website I published under.

Part 2.

2nd stone

Ek = 0.5mv^2 = 0.5m(47.04^2)

Ep = mgh = m(9.8)h

Ep = Ek

9.8h = 1106.38

h = 112.89m; height of the cliff

Part 1

1st stone

Ep = Ek

mgh = 0.5mv2

m(9.8)(112.89) = 0.5mv^2

1106.32 = 0.5v^2

v = 23.52; velocity when hits the ground

Ek = vt^2 = 23.52t^2

t = 4.85 seconds.