Question

A rocket is launched from atop a 104-foot cliff with an initial velocity of 110 ft/s. A. Substitute the values into the vertical motion formula . Let h = 0. B. Use the quadratic formula find out how long the rocket will take to hit the ground after it is launched. Round to the nearest tenth of a second.

Answer 1

Answer:

7.68 s

Step-by-step explanation:

The vertical motion formula is:

$h_{t} = h_{0} + v_{0}t - \frac{gt^{2}}{2}$

Where:

h(t): is the final height = 0

h(0): is the height of the cliff = 104 foot

v(0): is the initial velocity = 110 ft/s

g: is the gravity = 9.8 m/s² = 32.15 ft/s²

Hence, from the vertical motion formula we can find the time that will take the rocket to hit the ground:

$0 = 104 + 110*t - \frac{32.15t^{2}}{2}$

$0 = 104 + t(110 - 16.08t)$

We have two solutions:

t₁= -0.84 s

t₂ = 7.68 s

We will take the positive value, thus the time that will take the rocket to hit the ground is 7.68 s.

I hope it helps you!