Question

A rocket is launched at 85 ft./s from a launch pad that’s 28 feet above the ground. which equation can be used to determine the height of the rocket at a given time after the launch? (answer choices in picture)

Answer 1

The correct option is the last option

`h(t) = 28 + 85t -16t ^ 2`

Explanation:

The kinematic equation to calculate the position of a body on the vertical axis as a function of time is:

`h(t) = h_o + v_ot - (1)/(2)gt ^ 2`

Where:

`h_0` = initial position = 28ft

`v_0` = initial velocity =
`85\ (ft)/(s^2)`

g = acceleration of gravity =
`32.16\ (ft)/(s) ^ 2`

Then the equation sought is:

`h(t) = 28 + 85t - (1)/(2)32.16t ^ 2`

Finally:

`h(t) = 28 + 85t -16t ^ 2`

The correct option is the last option

Answer 2

`h(t)=-16t^2+85t+28` is equation of height of rocket.

Option D is correct.

Explanation:

Given: A rocket is launched with speed 85 ft/s from a height 28 feet.

Launching a rocket follows the path of parabola. The equation of rocket should be parabolic.

Parabolic equation of rocket is

Formula:
`h(t)=\frac{1}2gt^2+v_0t+h_0`

g ⇒ acceleration due to gravity (-32 ft/s)

v ⇒ Initial velocity (
`v_0=85\ ft/s`)

h ⇒ Initial height (
`h_0=28\ feet`)

h(t) ⇒ function of height at any time t

Substitute the given values into formula

`h(t)=(1)/(2)(-32)t^2+(85)t+28`

`h(t)=-16t^2+85t+28`

D is correct.