Rocket Challenge
Challenge: Launch a rocket and predict where it would would land. In order to do this, we had to use kinematic formulas to calculate and predict the horizontal distance traveled by the rocket. We initially set the rocket at 40° in order to calculate and predict the distance the rocket would travel at a greater launch angle.
Procedure: First, we collected data from a launch at 40°. We used a timer in order to know the time the rocket was traveling in the air and we used a contractor measuring wheel to find out the horizontal displacement of the rocket (in meters).
Diagram:
Diagram:
Data:
Time (s)
|
Distance
(m)
|
3.73
|
52.2
|
Calculations:
We then used the equation Vx= ∆x/t to solve for the horizontal velocity, Vx. Vx = 13.99 m/s. Using trigonometry, we can calculate cos(50°)= 13.99/vi
So, the actual velocity = 21.76 m/s
After we solved for the actual velocity, we now could apply this known velocity to our final test, since the launch velocity for the rocket is always the same.
We launched our rocket at a 25° angle for our test.
cos(65°)= vix / 21.76
vix = 9.2 m/s
cos(25°)= viy/ 21.76
viy = 19.72 m/s
∆x = 1/2at^2 + vit (∆xy = 0)
acceleration = -9.8 m/s^2 (acceleration due to gravity)
We can now plug in the known variables:
0 = -4.9t^2 + 19.72*t
Now we can use the quadratic formula:
t = -b + or - (√b^2-4ac)/(2a) Where a = -4.9, b = 19.72; c = 0
Solving the Equation: We calculated the time to be around 3.3 seconds. We then use Vx = ∆x / t (9.2 for Vx and 3.3 for t). Now we must solve for the horizontal Vx. We calculated the Vx (Vertical Displacement) to be 30.36 meters.
We set our target 30.36 meters from our launch site, and shot the rocket at an angle of 25°. Our rocket landed 32.3 meters from our launch site, which was exactly 1.94 meters off. So our percent error was 6.4%, which is higher than expected. We concluded that the wind most likely influenced the trajectory of the rocket.
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