SOLUTION: An object is launched with initial velocity of 22.5 meters per second (m/s) from a 75-meter tall platform. The equation for the object's height h at time t seconds after launched i

Question 389141: An object is launched with initial velocity of 22.5 meters per second (m/s) from a 75-meter tall platform. The equation for the object's height h at time t seconds after launched is h(t)=-4.9t^2+22.5t+75, where h is in meters. How long will it take for the object to reach the ground? what is the maximum height reached by the object?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
An object is launched with initial velocity of 22.5 meters per second (m/s) from a 75-meter tall platform. The equation for the object's height h at time t seconds after launched is h(t)=-4.9t^2+22.5t+75, where h is in meters. How long will it take for the object to reach the ground?
set h(t) to zero and solve for t:
0=-4.9t^2+22.5t+75
applying the quadratic formula, we get:
t = {-2.24, 6.83}
toss out the negative answer leaves:
t = 6.83 seconds
.
what is the maximum height reached by the object?
.
The vertex is the max height. It reaches this at:
t = -b/(2a)
t = -22.5/(2*(-4.9))
t = -22.5/(-9.8)
t = 2.30 seconds
.
h(2.30) = -4.9(2.30)^2+22.5(2.30)+75 = 100.83 meters (max height)

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