Question

A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 36t + 9. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?

A. 1.13 s; 69.75 ft


B.1.13 s; 29.25 ft


C.2.25 s; 9 ft


D.1.13 s; 31.5 ft

Answer 1

b is ur answer hope i helped

Answer 2

Answer: B) 1.13 s; 29.25 ft

Explanation:

Height of the ball h in feet after t seconds is given by the function

`h=-16t^(2)+36t+9`

Now at maximum height
`(\partial h)/(\partial t)=0</p><p>=>(\partial( -16t^(2)+36t+9))/(\partial t)=0`

=>
`\therefore -32t+36=0=>t=1.125\, seconds`

After rounding to nearest hundredth,
`t\simeq 1.13\, seconds`

Height ,
`h_(max)=-16t^(2)+36t+9`

`=>h_(max)=[-16(1.125)^(2)+36* 1.125+9] feet=29.25 feet`

Therefore , t=1.13 s and
`h_(max)=29.25 feet`

Thus correct option is B) 1.13 s; 29.25 ft