SOLUTION: A ball is thrown upward with an initial velocity of 30 feet per second from a point that is 24 feet above the ground. The height (h) in feet of the ball at time t (in second) is gi
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Question 1165980: A ball is thrown upward with an initial velocity of 30 feet per second from a point that is 24 feet above the ground. The height (h) in feet of the ball at time t (in second) is given by the equation: h(t)= -16t^2+30t+24
a) at what time did the ball reach its maximum height?
b) what was the maximum height?
Thanks!! Answer by ikleyn(52788) (Show Source):
The function h(t) = -16t^2 + 30t + 24 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= 30, so the function gets the maximum at t = = 0.9375.
So, the ball gets the maximum height 0.9375 seconds after is hit straight up.
The maximum height then is h(0.9375) = -16*0.9375^2 + 30*0.9375 + 24 = 38.0625 ft.
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