# Velocity of an object dropped

1. May 1, 2013

### Dustobusto

1. The problem statement, all variables and given/known data

What is the velocity of an object dropped from a height of 300 m when it hits the ground?

2. Relevant equations

s(t) = s0 +v0t -1/2gt2

v(t) = v0 - gt

s0 = initial height

v0 = initial velocity

v(t) = s'(t)

3. The attempt at a solution

Attempted this for a while, can't seem to get it right. I know initial height = s0 = 300m.

g, as given by the book = 9.8 (represents gravity)

I understand that when trying to find maximum height, velocity = zero. For this, I assume maximum velocity is found when s(t) = zero. Should I believe that when s0 = 300, v0 and t = zero?

2. May 1, 2013

### EngnrMatt

Remember that g is -9.81 m/s2. That is, it is a negative value.
try this equation:

v2 = v02 + 2g(x-x0)

3. May 1, 2013

### Dustobusto

I appreciate you giving me a new equation, but I'm trying to solve this within the context of the information given. I doubt I'm supposed to solve using that equation. No offense intended.

4. May 1, 2013

### nil1996

From the given eqn.s
you can use s(t) = s0 +v0t -1/2gt2
put s0 =0;v0t=0;g=9.8m/s2;s(t)=300m

from this you will get time(t) required for the body to reach the ground.

now take the equation

v(t) = v0 - gt

put v0 =0;g=9.8m/s2
also put the time(t) found from the previous eqn.

v(t) is the required velocity.

5. May 1, 2013

### Dustobusto

Dammit. Thanks nil. Dammit dammit..I was so close. I just had a few things wrong here and there..

But yeah, I had the right idea, was so close. Oh well.