Question

Suppose you launch a model rocket with an upward starting velocity of v ft/s. You can use the equation to find the rocket's altitude, h represents height in feet, t seconds after launch and represents initial height. Suppose the upward starting velocity is 315 ft/s and the initial height is 3 ft. When will the rocket hit the ground?

Answer 1

Answer:

The rocket will hit the ground at 64.2 seconds.

Step-by-step explanation:

The time at which the rocket will hit the ground can be calculated using the following equation:

$Y_{f} = Y_{0} + V_{0}y*t - \frac{1}{2}gt^{2}$

Where:

$Y_{f}$: is the final height = 0

$Y_{0}$: is the initial height = 3 ft

$V_{0}$: is the initial speed = 315 ft/s

t: is the time =?

g: is the gravity = 9.81 m/s²

$3 + 315*t - \frac{1}{2}9.81*t^{2} = 0$        

Solving the above quadratic equation for t, we have the following values:

t₁ = -0.0095 s

t₂ = 64.2 s

Taking the positive value, we have that the rocket will hit the ground at 64.2 seconds.

I hope it helps you!