SOLUTION: An object is thrown straight up from the roof's edge of a 48-foot building at an initial velocity of 32 ft/s. How long will it take to reach the ground?
Question 1051153: An object is thrown straight up from the roof's edge of a 48-foot building at an initial velocity of 32 ft/s. How long will it take to reach the ground? Answer by ikleyn(52817) (Show Source):
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An object is thrown straight up from the roof's edge of a 48-foot building at an initial velocity of 32 ft/s.
How long will it take to reach the ground?
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The governing equation is
h(t) = -16t^2 + 32t + 48,
where t is the time after throwing, h(t) is the object height (elevation over the ground surface in feet,
32 is the initial velocity in ft/s, 48 is the initial elevation, 16 = a is the half of the value
of the free fall acceleration in ft/s^2.
We need to find the value of t for which h(t) = 0, or
-16t^2 + 32t + 48 = 0.
Divide both sides by (-16). You will get
t^2 - 2t - 3 = 0.
Factor the left side
(t-3)*(t+1) = 0.
The positive root t = 3 is the time we are seeking for.
The other root is negative, so we ignore it.
Answer. It will take 3 seconds for the onject to reach the ground.
