Question

A ball is thrown directly upward from a height of 5 ft with an initial velocity of 24 ​ft/sec. The function ​s(t)=−16t2+24t+5 gives the height of the​ ball, in​ feet, t seconds after it has been thrown. Determine the time at which the ball reaches its maximum height and find the maximum height.

Answer 1

The ball reaches 14 ft in 0.75 seconds

Explanation:

`h(t) = -16t^2+24t+5\\h'(t) = -32t+24\\`

The highest point occurs when the derivative is equal to 0, because the tangent line is horizontal at that point

`-32t + 24 = 0\\24 = 32t\\t = (24)/(32)=(3)/(4)=0.75s\\`

plug t back into h(t) to get height

`h(0.75) = -16(0.75)^2+24(0.75)+5\\=(-16)(9/16)+23\\=-9 + 23\\= 14 ft`

Note: If you are not familiar with the concept of derivatives yet, you can find the time of the highest point using the equation
`t =-(b)/(2a)` where
`h(t) = at^2 + bt + c`