The Height H(t) (in feet) of a projectile fired vertically into the air from the ground is given by the equation H(T) = -16t^2 + 96t, where t is time (in sec) after the launch. A) How long will it take the projectile to reach its maximum height?

B) What is the maximum height?

C) How long does it take the projectile to reach the ground again.

a) 3s

B) 144m

C) 6s

Step-by-step explanation:

Given the height modeled by the function;

H(T) = -16t^2 + 96t,

The body will reach its maximum height when dh/dt = 0 (when velocity is zero)

dh/dt = -32t + 96

0 = -32t + 96

32t = 96

t = 96/32

t = 3s

Hence it will take the projectile 3s to reach its maximum height'

B) To get the maximum height we will substitute t = 3 into the equation

H(T) = -16t^2 + 96t,

H(3) = -16(3)^2 + 96(3)

H(3) = -144+ 288

H(3) = 144m]

Hence the maximum height is 144m

C) The projectile reaches the ground at h = 0

H(T) = -16t^2 + 96t,

0= -16t^2 + 96t,

16t² = 96t

16t = 96

t = 96/16

t = 6

Hence it will take the projectile 6s to reach the ground again

tube or goog would be a great resource to with that question

75

step-by-step explanation:

you do x-29=46 (the number of cantaloupes sold)

in order to find x, you have to find what minus 29 equals 46, meaning you'd have to add 29 to 46.

29+46=75

if you replace x with 75, the equation looks like:

75-29=46

x=75