Question 1177200: A ball is thrown vertically downward from the top of a building, leaving the thrower’s hand with a velocity
of 30 ft/s.
(a) What will be its velocity after falling for 2 sec?
Found 2 solutions by Solver92311, ikleyn:
Answer by Solver92311(821)   (Show Source):
You can put this solution on YOUR website!
Your units are feet and seconds, so distance in feet, time in seconds, the magnitude of velocity is feet per second, and magnitude of the acceleration is feet per second per second.
The presumption here is that the building in the story is on the surface of planet Earth where the acceleration due to gravity is  and motion downward, i.e. toward the center of the earth, is indicated by negative acceleration and velocity magnitudes.
The height as a function of time for an object traveling vertically with respect to the earth's surface is:
\ =\ \frac{1}{2}gt^2\ +\ v_ot\ +\ h_o)
Where  is the acceleration due to the force of gravity,  is the initial velocity magnitude (recall that the velocity magnitude must be a negative quantity for an object thrown downward), and  is the initial height.
The instantaneous velocity as a function of time for this situation is the first derivative of the height function, namely:
So
\ -\ 30\ \text{ft/sec})
You can do your own arithmetic.
John
My calculator said it, I believe it, that settles it
From
I > Ø
Answer by ikleyn(52786)  (Show Source):
You can put this solution on YOUR website! .
From Physics, in this problem vertical velocity in 2 seconds is
v =  - gt = -30 ft/s - 32 ft/s^2 * 2 seconds = -30 - 64 = - 94 ft/s.
directed vertically down.
Solved, answered, explained and completed.
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