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?

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#