SOLUTION: you are on a roof of a building. you throw a ball into the air with a velocity of 32ft/s. the ball is 48 ft above the ground when it leaves your hand. how many seconds will it take

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Question 317352: you are on a roof of a building. you throw a ball into the air with a velocity of 32ft/s. the ball is 48 ft above the ground when it leaves your hand. how many seconds will it take for the ball to reach the ground? use the vertical motion formula -16t^+ vt + s =h.
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
You are on a roof of a building. You throw a ball into the air with a velocity of 32ft/s. the ball is 48 ft above the ground when it leaves your hand. how many seconds will it take for the ball to reach the ground? use the vertical motion formula -16t^+ vt + s =h.
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Height of the ball is zero when it hits the ground.
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-16t^2 + 32t + 48 = 0
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Divide thru by -16 to get:
t^2 - 2t - 3 = 0
(Factor:
(t-3)(t+1) = 0
Positive Solution:
t = 3 seconds
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Cheers,
Stan H.
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Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
you are on a roof of a building. you throw a ball into the air with a velocity of 32ft/s. the ball is 48 ft above the ground when it leaves your hand. how many seconds will it take for the ball to reach the ground? use the vertical motion formula -16t^+ vt + s =h.
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h(t) = -16t^2 + vt + s
v = initial velocity = 32 ft/sec
s = initial height = 48 feet
h(t) = -16t^2 + 32t + 48
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Find when h(t) = 0
-16t^2 + 32t + 48 = 0
-t^2 + 2t + 3 = 0
(-t - 1)*(t - 3) = 0
t = -1 second (Ignore)
t = 3 seconds