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(52792)   (Show Source): You can put this solution on YOUR website!
.

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.
ANSWER

Solved.

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On finding the maximum/minimum of a quadratic function,  see the lessons
    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola
in this site.

On solving similar problems to yours in this post,  see the lessons
    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform
in this site.


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