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

The rocket will hit the ground at 64.2 seconds.

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 - (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 - (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!