Question

The height of a basketball `t` seconds after it is thrown can be modeled by the function `y=-16t^2 + 32t + 2`. What is maximum height of the basketball? How long is the basketball in the air, if its caught in its descent 7 feet above the ground?

Answer 1

Maximum height is `y=18 ft` and time in air is `t~=1.83 sec`

Explanation:

graph{-16x^2 + 32x + 2 [-20.67, 23.72, -1.34, 20.86]}
Consider this graph where height (y) is on y axis and time (t) is on x axis
You can imagine that at maximum height basketball's velocity becomes zero ,so it sort of hovers in air and does not go up or down
for a very short time.

So at maximum height slope of `y=>t` graph becomes zero

For given equation differentiating with respect to time we get velocty

`y=-16t^2 + 32t + 2`

`(dely)/(delt)=-32t+32`

`(dely)/(delt)` is velocity which at maximum height is zero

`(dely)/(delt)=-32t+32 =0`

`-32t=-32`

`t=1`

Hence at `t=1` second ball is at maximum height which makes sense if you look at graph given above,at `t=1` second there is no change in height so it is hovering

Max height(H)`=-16(1)^2+32(1)+2` `=18`

It takes 1 second for the ball to reach maximum height and then it starts its descent

If we plug `y=7` in our equation we should expect two values of `t` because ball is at `y=7` two times namely ascent and descent but we want its time during its descent hence we will take the time of larger magnitude as our answer

`7=-16t^2+32t+2`

`0=-16+32t-5`

We get `t_1=1 - sqrt(11)/4` and `t_2=1 + sqrt(11)/4`

But we will take `t_2` as our answer

`t_2~=1.83`