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:
= time (I'll assume it's in seconds)
= initial height
= initial velocity
= height after t sec
= velocity after t sec
The equation you've given us is
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?
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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