SOLUTION: An object is shot straight up into the air from ground level with initial speed of 24 ft/sec. The height of the object (in feet) is given by the equation h=-16tsquared+24t, where t
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Question 79507: An object is shot straight up into the air from ground level with initial speed of 24 ft/sec. The height of the object (in feet) is given by the equation h=-16tsquared+24t, where t is the time in seconds after launch. Find the time(s) when the object is at ground level.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
An object is shot straight up into the air from ground level with initial speed of 24 ft/sec. The height of the object (in feet) is given by the equation h=-16tsquared+24t, where t is the time in seconds after launch. Find the time(s) when the object is at ground level.
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At ground level the height is zero.
To get the time let h=0 and solve for t:
-16t^2+24t=0
-8(2t^2-3t)=0
2t^2-3t=0
t(2t-3)=0
t=0 or t=3/2
This means the object was on the ground to start with (t=0)
and will be on the ground again in 1.5 seconds (t=3/2)
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Cheers,
Stan H.
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