Aprojectile that was launched straight up from the ground with an initial velocity of 48 ft/s returned to the ground after 3 s. the height of the projectile t seconds after launch is modeled by the function f(t)=−16t2+48t . what is the maximum height of the projectile, in feet? enter your answer in the box.

At the maximum height, the velocity of the projectile would be zero. We can calculate the maximum height by taking the first derivative of the function given and equating it to zero to obtain the time it take to reach the max height. We do as follows:

f(t)=−16t2+48t
f'(t) = -32t + 48 = 0
t = 1.5 s

max height =−16(1.5)^2+48(1.5) = 36 m

The linear equation is: y=ax+b