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

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p

find the vertical and horizontal asymptotes of the function f x = 5 x − 1 . to find the vertical asymptote, equate the denominator to zero and solve for . since the degree of the polynomial in the numerator is less than that of the denominator, the horizontal asymptote is .

step-by-step explanation: