The path of a projectile launched from a 16-ft-tall tower is modeled by the equation y = −16t2 + 64t +16. What is the maximum height, in feet, reached by the projectile?

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

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step-by-step explanation:

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