# A stone is thrown straight up from the edge of a roof, 1000 feet above the ground, at a speed of 14 feet per second. What is the velocity of the stone when it hits the ground?

The equation for displacement is `s(t)=-16t^2+v_0t+s_0` where `v_0` isthe initial velocity and `s_0` is the initial displacement. We adopt the convention that velocity towards the earth is negative:

So we have `s(t)=-16t^2+14t+1000` The velocity function is the first derivative of the displacement function with respect to t:

`v(t)=-32t+14`

Solving for when the stone hits the ground we get:

`t=(-14+-sqrt(14^2-4(-16)(1000)))/(-32)` ==>`t~~-7.48,t~~8.355` seconds.

So with `t~~8.355` seconds, we find the velocity to be :

`v(8.355)=-32(8.355)+14=-253.37` `"ft"/"sec"`

The velocity is approximately 253.4 feet per second towards the ground.

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