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+28 28tplus+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.

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:

The function s(t) reaches its maximum height at the axis of symmetry. For a quadratic equation of the form , the axis of symmetry occurs at .

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

The ball reaches its maximum height after 0.875 seconds

(b)Maximum Height

Given

At t=0.875

Maximum height = 78.25 ft

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

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent

answer: $1794

step-by-step explanation: