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!
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?
$\displaystyle s(t) = -6.1t^2 + 75$
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.Or do you just plug in the 3 for t to get 54.9 ft/sec?
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