SOLUTION: a rocket is launched from the ground with an initial velocity of 48 feet per second, so that its distance in feet above the ground after t seconds is s(t)=-16t^2+48t. what is the m
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Question 1189385: a rocket is launched from the ground with an initial velocity of 48 feet per second, so that its distance in feet above the ground after t seconds is s(t)=-16t^2+48t. what is the maximum height of the rocket Found 2 solutions by ikleyn, Alan3354:Answer by ikleyn(52794) (Show Source):
You can put this solution on YOUR website! .
a rocket is launched from the ground with an initial velocity of 48 feet per second,
so that its distance in feet above the ground after t seconds is s(t)=-16t^2+48t.
what is the maximum height of the rocket
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The function s(t) = -16t^2 + 48t is a quadratic function, whose plot is a parabola opened down.
This quadratic function has the maximum at the value of its argument t = , where
"a" is the coefficient at t^2 and "b" is the coefficient at t.
In your case, a= -16, b= 48, so the function gets the maximum at t = = 1.5 seconds.
So, the ball gets the maximum height 1.5 seconds after is hit straight up.
The maximum height is then s(1.5) = - 16*1.5^2 + 48*1.5 = 36 feet. ANSWER
