Ryan ThomasA.P. Physics, Per. 5/69/30/07

Rocket Lab Write-Up

Purpose: To practice calculations of projectile motion in an entertaining and interactive way, through attempting to hit a stationary target with rockets.

Equipment: Rocket (Nose Cap and Shaft) Firing Cap, Medium size Pump (Launcher) Wedge (For creating angles) Protractor (For measuring wedge)

Assumptions:In the following calculations, there are two main assumptions. First: air resistance

does not exist. This means that the distance calculated for the rocket should be greater than the actual distance it will travel, but by so little as for the difference to be negligible. Second: wind is not a factor. This assumption allows for simple calculation; we do not have to calculate how much or little the wind blows, in what directions, and how it would affect the rocket’s flight.

Calculations: Known Information: A rocket launched straight up, using the medium firing cap, took an average time of 4.5 seconds to return to the ground; The angle used to fire our rocket will be 30º, due to the

Problem: How must the same rocket be aimed to hit a target with a height of 1.21 meters; that is, at what distance from the target must the rocket be fired to hit Mr. Balantic head-on?

Step 1 – Find υor

Simple enough; just plug the numbers from the straight-up launch into the height equation:

0 = 0 + υor(4.5) + -4.9(4.5)2 → υor = 22.05 m/s

Step 2 – Find time of height 1.21 meters

Again, fairly straightforward. Using the newly found υor, recognize that the rocket’s initial velocity in the vertical = υorsinθ = 22.05sin30 m/s (see fig 1). Then, plug this into the height equation of the rocket, to find how long it takes to reach a height of 1.21 meters. Two times result, as the equation becomes a quadratic; the later one is correct, as this is when the rocket is landing, and the previous time is for when the rocket reaches 1.21 meters as it launches.

1.21 = 0 + 22.05sin30(t) + -4.9(t)2

_______________11.025 +/- √(11.025) 2 – 23.716 9.8

t = 2.13 or 1.34

Step 3 – Find necessary distance

Now that the time of the rocket’s necessary travel is known, plug it into the distance equation; υorcosθ = 22.05cos30 m/s = The rocket’s velocity in the horizontal

dr = 2.13s * 22.05cos30 m/s = 40.76m

Also, since we’re on a football field, let’s convert that to yards:

1 meter = 1.0936133 yard; 40.76m * 1.0936133 yards/1 meter ≈ 44.6 yards

= t