SOLUTION: If you were to throw an object straight up from a platform 50 feet high, at an initial velocity of 80 feet per second, the height of that object in feet above the ground t seconds

Question 93610: If you were to throw an object straight up from a platform 50 feet high, at an initial velocity of 80 feet per second, the height of that object in feet above the ground t seconds after you released it would be described as follows:
Height = -16*t^2 + 80*t + 50
How high will tge object be 3.5 seconds after released?

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
If you were to throw an object straight up from a platform 50 feet high, at an initial velocity of 80 feet per second, the height of that object in feet above the ground t seconds after you released it would be described as follows:
:
Height = -16*t^2 + 80*t + 50
:
How high will the object be 3.5 seconds after released?
:
Substitute 3.5 for t and find the height:
h = -16(3.5^2) + 80(3.5) + 50
h = -16(12.25) + 280 + 50
h = -196 + 280 + 50
h = 134 ft high in 3.5 sec
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