Question 1181591: a ball is launched upwards from the edge of a cliff. the equation for the object’s height h(t), at the time, (t seconds) after launch is h(t)= -5t^3 + 20t + 5 where h(t) is in meters. what is the maximum height of the ball and what time does it reach this height? at what time is the ball 10 meters off of the ground, round the answer to the nearest tenth of a second.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! h(t)=-5t^2+20t+5, editing the question to t^2 not t^3 because the gravitational force problems are quadratic, not cubic, and the metric coefficient is -4.9, which is sometimes rounded to 5.
max height is at vertex which has x=-b/2a or -20/-10 or at 2 sec. That is the time required.'
h(2)=-20+40+5=25 feet
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at 10 m off the ground, h(t)=10. There are two solutions, one going up, the other coming down.
so -5t^2+20t-5=0
-5(t^2-4t+1)=0
t=(1/2)(4+/- sqrt 16-4))
t=(1/2)(4+/-2 sqrt(3)
=2+/- sqrt (3)
this is at 0.268 sec and 3.732 sec or 0.3 and 3.7 sec.
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