A model rocket is shot into the air from ground level with the upward velocity of 80 feet per second. Find the time taken by the rocket to reach the height of 96 feet.

## Story problem

# 1 Answer

This type of motion is described as projectile motion, ideally in one dimension meaning that the rocket is going straight up and straight down.

The equation for motion in one direction is given by

d = v_{i}t + (1/2)at^{2}

where d is the displacement

v_{i} = initial velocity

a = acceleration

t = time

In this case though the only acceleration is on the rocket due to gravity and is equal to 32 ft/s^{2}. Also, because the acceleration is opposed to the direction of the initial velocity it is given a negative sign.

So, plugging your values into the equation you get:

**96 = 80t - (1/2)*32t ^{2}**

Rewriting that yields

**16t ^{2} - 80t + 96 = 0**

There's your quadratic equation. As it turns out, all of those coefficients are divisible by 16 so it simplifies:

**t ^{2} - 5t + 6 = 0**

Factoring that gives you two solutions since

**(t - 2)(t - 3) = 0**

**then at both
**

** t = 2 seconds
**

**and t = 3 seconds
**

**t**he rocket is at a height of 96 ft. Both times satisfy the equation because the path of travel is parabolic. The rocket rises to a height of 96 ft, continues traveling for another half second until gravity slows its upward velocity to zero,
and then begins to fall back toward the ground reaching 96 ft again at 3 seconds into the flight.