Question

A ball is thrown directly upward from a height of 66 ft with an initial velocity of 2828 ​ft/sec. The function ​s(t)equals=minus−16tsquared2plus+2828tplus+66 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

Answer:

time: 0.875 seconds

maximum height: 18.25 feet

Step-by-step explanation:

The function that gives the height of the ball in feet after t seconds is:

s(t) = −16t^2 + 28t + 6

The inicial velocity is 28 ft/sec, and the inicial position is 6 feet.

So, to find the time when the ball reaches the maximum heigth, we need to find the vertix of the equation −16t^2 + 28t + 6.

We can use the following formula to find it:

t_vertix = -b/2a

where a and b are coefficients of the quadratic equation (in our case, a = -16 and b = 28).

So, we have that:

t_vertix = -28/(-32) = 0.875 seconds

To calculate the maximum height we just need to use this time in the equation of position:

s(0.875) = −16*(0.875)^2 + 28*0.875 + 6 =  18.25 feet

Answer 2

Answer:

Time taken to reach maximum height=0.875 seconds

Maximum height = 78.25 ft

Step-by-step explanation:

The function which models the height of the ball is given as:$s(t)=-16t^2+28t+66$

The function s(t) reaches its maximum height at the axis of symmetry. For a quadratic equation of the form $ax^2+bx+c=0$, the axis of symmetry occurs at $x=-\frac{b}{2a}$.

In s(t), a=-16, b=28

$t=-\frac{28}{2(-16)}=0.875 seconds$

The ball reaches its maximum height after 0.875 seconds

(b)Maximum Height

Given $s(t)=-16t^2+28t+66$

At t=0.875

$s(0.875)=-16(0.875)^2+28(0.875)+66=78.25 ft$

Maximum height = 78.25 ft