# Free fall

• June 9th 2009, 04:41 PM
live_laugh_luv27
Free 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!
• June 9th 2009, 05:12 PM
sa-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.
• June 9th 2009, 05:48 PM
live_laugh_luv27
ok..then how would I set this problem up?
• June 9th 2009, 08:51 PM
yeongil
They are a lot of things wrong with the original post. First, let's look at the vertical motion equations on Earth:
$s(t) = -16t^2 + v_0 t + s_0$
$v(t) = -32t + v_0$

$t$ = time (I'll assume it's in seconds)
$s_0$ = initial height
$v_0$ = initial velocity
$s(t)$ = height after t sec
$v(t)$ = velocity after t sec

The equation you've given us is
$s = 6.1t^2$
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?
$s(t) = -6.1t^2 + 75$

Quote:

Or do you just plug in the 3 for t to get 54.9 ft/sec?
Nope. The equation above looks like a position equation, not a velocity equation. Using the equation as originally given, plugging 3 in for t gives us a height of 54.9 feet, NOT a velocity of 54.9 ft/sec.

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