Find the maximum height of an object launched with an initial vertical velocity of 32 feet per second and an initial height of 4 feet.

Answer 1

Quadratic formula of a projectile in motion (lunched upward):

s_(t) = -g t² + v₀ t + h₀

Where:

s_(t): position at time t

-g t²: acceleration = -4.9t²

v₀ t: initial velocity

h₀: initial height

In this case:

s_(t) = -4.9t² + 32t + 4

Maximum height in a negative quadratic, is the vertex of the parabola:

t = -b / 2a

t = -32 / 2 (-4.9)

t = -32 / -9.8

t = 3.27

s_(3.27) = -4.9 (3.27)² + 32 (3.27) + 4

s_(3.27) = 56.24

The maximum height is 56.24 feet, at 3.27 seconds.

Answer 2

h(t) =-16t^2 +32v + 4

take the derivative and set it =0, solve for t

h'(t) = -32t +32 = 0, t=1 at max height

plug t=1 into the original h(t) equation to get

h(1) = -16(1)^2 +32(1) + 4 = 36 feet = max height