SOLUTION: A projectile is launched upward at an initial velocity of 64 ft/sec from a platform. It took 8 sec to hit ground. How tall was the platform?

Question 101844: A projectile is launched upward at an initial velocity of 64 ft/sec from a platform. It took 8 sec to hit ground. How tall was the platform?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A projectile is launched upward at an initial velocity of 64 ft/sec from a platform. It took 8 sec to hit ground. How tall was the platform?
:
The height of the platform will be the 3rd term in the equation (call it k),
Where x = time in seconds and h = height of the projectile
:
h = -16x^2 + 64x + k
:
Height of the ground is 0 so we can write it:
-16x^2 + 64x + k = 0
:
It told us it hit the ground in 8 seconds, substitute 8 for x
-16(8^2) + 64(8) + k = 0
-15(64) + 512 + k = 0
-1024 + 512 + k = 0
k - 512 = 0
:
k = 512 ft is the height of the platform
:
:
You can confirm this by finding x using the completed equation
-16x^2 + 64x + 512 = 0
Simplify divide by -16
x^2 - 4x - 32 = 0
Factors to:
(x-8)(x+4) = 0
x = +8 seconds confirms our solution
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