Equation given -- s=6.1t^2, t is seconds.

Object dropped from 75 feet. Find the speed of the object at t = 3.

one question- is the 75 feet important? Or do you just plug in the 3 for t to get 54.9 ft/sec?

Any help is appreciated!

Printable View

- Jun 9th 2009, 04:41 PMlive_laugh_luv27Free fall
Equation given -- s=6.1t^2, t is seconds.

Object dropped from 75 feet. Find the speed of the object at t = 3.

one question- is the 75 feet important? Or do you just plug in the 3 for t to get 54.9 ft/sec?

Any help is appreciated! - Jun 9th 2009, 05:12 PMsa-ri-ga-ma
In the problem speed of the object is required.

Speed v = ds/dt.

In the problem 75 feet is necessary. Otherwise the object may reach the ground before 3 seconds. - Jun 9th 2009, 05:48 PMlive_laugh_luv27
ok..then how would I set this problem up?

- Jun 9th 2009, 08:51 PMyeongil
They are a lot of things wrong with the original post. First, let's look at the vertical motion equations on Earth:

$\displaystyle s(t) = -16t^2 + v_0 t + s_0$

$\displaystyle v(t) = -32t + v_0$

$\displaystyle t$ = time (I'll assume it's in seconds)

$\displaystyle s_0$ = initial height

$\displaystyle v_0$ = initial velocity

$\displaystyle s(t)$ = height after t sec

$\displaystyle v(t)$ = velocity after t sec

The equation you've given us is

$\displaystyle s = 6.1t^2$

Does up => positive and down => negative? Or the other way around? If it is the former, then this equation is for an object that floats upwards. (In other words, is there a negative sign missing?)

Since the object is being dropped, the initial velocity is zero, so it's okay that there is no t term in the equation you gave us. But there is mention of the initial height at all. So the question is, is this the equation you meant to give us?

$\displaystyle s(t) = -6.1t^2 + 75$

Quote:

Or do you just plug in the 3 for t to get 54.9 ft/sec?

**feet**, NOT a velocity of 54.9 ft/sec.

01