Find the Initial Height of a Falling Block

[c4iscool]

Where do I start?
A 10kg block being held at rest above the ground is released. The block begins to fall under only the effect of gravity. At the instant that the block is 2.0 meters above the ground, the speed of the block is 2.5m/sec. The block was initially released at a height of how many meters.

 

[neutrino]

Hint: Energy conservation.

 

[c4iscool]

is that the equation m1vf^2-1/2mvi^2, if it is how would u find the height it was released at? or are you talking about potential energy?

 

[neutrino]

is that the equation m1vf^2-1/2mvi^2, if it is how would u find the height it was released at?

It's mgh_1 + \frac{1}{2}mv_1^2 = mgh_2 + \frac{1}{2}mv_2^2

The left side (initial) variables are at the instant when the block was at the "top", and those on the right (final) are the instant when the block is at the given height of 2m. Therefore, it is Initial energy = Final energy.

 

[c4iscool]

so your saying:
(10kg)(9.8m/sec)h+(1/2)10kg(0^2)=10kg(9.8m/sec)(2m)+(1/2)10kg(2.5m/sec)^2

 

[c4iscool]

i should have it from here. thanks

 

[neutrino]

10\times 9.8 h = 10\times 9.8 (h - 2) + \frac{1}{2}\times 10 \times 2.5^{2}

Correct me if I'm wrong, Hootenanny, but isn't that the expression for the block which has fallen a distance of two metres from the initial position. In fact if you expand the terms, you get 196 = 31.25.

 

[Hootenanny]

Correct me if I'm wrong, Hootenanny, but isn't that the expression for the block which has fallen a distance of two metres from the initial position. In fact if you expand the terms, you get 196 = 31.25.

My bad. I apologise the OP had it correct first time round. . The only excuse I can offer is that it was rather late last night when I posted . Thanks for pointing that out neutrino