Answer:
80 feet
Step-by-step explanation:
Given the path of a projectile launched from a 16-ft-tall tower modeled by the equation y = −16t^2 + 64t +16
h is the maximum height reached
t is in seconds
The velocity of the projectile at its maximum height is zero;
velocity v = dy/dt
v = -32t + 64
Since v = 0 at maximum height
0 = -32t + 64
32t = 64
t = 2secs
Substitute t = 2secs into the equation of the height to get the maximum height.
y = −16t^2 + 64t +16
at t = 2secs
y = −16(2)^2 + 64(2) +16
y = -64 + 128 + 16
y = 64+16
y = 80 feet
Hence the maximum height, in feet, reached by the projectile is 80 feet
