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

answer:  

123455421

step-by-step explanation: easy

2. angles qrt and qst both measures 156 degrees.

step-by-step explanation:  

we have been given a diagram of a circle and we are told that measure of arc qvt is 156 degrees.  

we can see from our given diagram that angle qrt and angle qst subtends to arc qvt.

since we know that angles subtended by the same arc at the circumference are always equal.

by angle subtended by same arc theorem, measure of angle qrt and angle qst will be equal to the measure of arc qvt.

therefore, angles qrt and qst both measures 156 degrees and 2nd statement is the correct choice.