A pistol is accidentally discharged vertically upward from a height of 3 feet above the ground. If the bullet has an initial velocity of 200 feet per second the path of the bullet is modeled by the equation h=-16t 2 +200t+3, what maximum height will it reach before it starts to fall back down to the ground?

hmax = 628 ft

Step-by-step explanation:

The bullet will reach its maximum height when its velocity h'(t) becomes zero. The equation of motion is

h(t) = -16t^2 + 200t + 3

Taking the derivative of h(t) is

h'(t) = -32t + 200

Solving for t when h'(t) = 0, we get

t = 200/32 = 6.25 s

and this is the time it takes for the bullet to reach its maximum height. Using this value for t into h(t), we get

h = -16(6.25 s)^2 + 200(6.25 s) + 3

= 628 ft

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.