# Free fall

1. Sep 29, 2004

### elsmith2

need help finding where to start,

"A freely falling object requires 1.50sec to travel the last 30.0m before it hits the ground. From what height above the ground did it fall?"

I know that:
accel = -9.80m/s^2
avg. veloc for last 1.50 sec = 20m/s
that i need the initial velocity for that period

Ed

2. Sep 29, 2004

### elsmith2

sitting here looking at my post and my physics book it dawned on me to use this eqn:
() = subscript
x(f) = x(i) + v(xi)t + 1/2 a(x)t^2

therefore

30 = 0 +v(1.50) + 1/2(-9.80)(1.50^2)
v = 27.35m/s

but that cant be, so i switched the accel to positive and got v = 12.65m/s which sounds more reasonable, but i thought that we use -g for accel of free falling objects. now i'm confused.

3. Sep 29, 2004

### vsage

ok let's do a little modification to the displacement formula

originally we have d = d1 + v1*t + 0.5*a*t^2
Since it's free falling let's just make v1 = 0 and assume it was just dropped from a height d1

d = d1 + 0.5*a*t^2 where d1 was the original height from which it began a descent

let t' = t - 1.50 so then d' would be the displacement at time t' (30 m above the ground). I'm using t as the total time it took for the descent.

d' = d1 + 0.5*a*t'^2

30 = d1 - 0.5 * 9.8*(t-1.5)^2 Equation (1)

You also know that
0 = d1 + 0.5*a*t^2
0 = d1 - 0.5 * 9.8*(t)^2
d1 = 0.5 * 9.8 * (t)^2 Equation (2)

substitute d1 back in to equation 1 to get

30 = 0.5*9.8*t^2 - 0.5*9.8*(t-1.5)^2

solve for t. you should get around 2.8 seconds.

so now you know d1 = -0.5*a*t^2 by equation 2 so plug in t and you're done!
I could have done this a lot easier but I wanted to show you step by step.

Last edited by a moderator: Sep 29, 2004