Question

An amateur rocket club is holding a competition. They are launching rockets from the ground with an initial velocity of 315 ft/sec. There is a cloud cover at an altitude of 1000 feet.
1: How long will it take until the rocket is out of sight?

2. How long will the rocket be in the air?

Answer 1

Problem 1

The projectile formula is

h = -16t^2 + vt + s

where,

• t = time in seconds
• h = height at time t
• v = initial or starting velocity
• s = starting height

In this case, we're starting from the ground so s = 0. The starting velocity is v = 315. This formula only works if you're in feet. If you work with meters, then you'll need a slightly different formula.

Plug s = 0 and v = 315 into the equation to get

h = -16t^2 + vt + s

h = -16t^2 + 315t + 0

h = -16t^2 + 315t

Next, replace h with 1000. We'll solve for t so we can find out when the rocket will reach this height.

h = -16t^2 + 315t

1000 = -16t^2 + 315t

0 = -16t^2 + 315t - 1000

-16t^2 + 315t - 1000 = 0

Let's use the quadratic formula to solve for t.

$t = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\t = \frac{-(315)\pm\sqrt{(315)^2-4(-16)(-1000)}}{2(-16)}\\\\t = \frac{-315\pm\sqrt{35225}}{-32}\\\\t \approx \frac{-315\pm187.68324379}{-32}\\\\t \approx \frac{-315+187.68324379}{-32} \ \text{ or } \ t \approx \frac{-315-187.68324379}{-32}\\\\t \approx \frac{-127.31675619}{-32} \ \text{ or } \ t \approx \frac{-502.68324379}{-32}\\\\t \approx 3.97864863 \ \text{ or } \ t \approx 15.70885137\\\\t \approx 3.98 \ \text{ or } \ t \approx 15.71$

This tells us two things:

1. The rocket enters the cloud cover at around 3.98 seconds
2. The rocket falls back down exiting the clouds at around 15.71 seconds

In other words, the timespan between approximately 3.98 seconds and 15.71 seconds is when the rocket is in the clouds and not visible. Outside this time span the rocket is visible.

We'll only focus on the smaller t value because your teacher is only worried about how long it takes for the rocket to get concealed by the cloud.

Answer: Approximately 3.98 seconds

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Problem 2

We'll return to this equation

h = -16t^2 + 315t

This time plug in h = 0 to find out when the rocket has hit the ground.

h = -16t^2 + 315t

0 = -16t^2 + 315t

-16t^2 + 315t = 0

t(-16t + 315) = 0

t = 0 or -16t + 315 = 0

t = 0 or -16t = -315

t = 0 or t = -315/(-16)

t = 0 or t = 19.6875

Ignore t = 0 because that's the rocket's initial time value. We have the rocket start on the ground, so of course this makes sense to be a solution.

The other solution is what we're after. At exactly 19.6875 seconds, the rocket will hit the ground. This is the timespan that the rocket is in the air.

Answer: Exactly 19.6875 seconds