Projectile Motion Lab Report

Topics: Probability theory, Expected value, Random variable Pages: 2 (559 words) Published: December 10, 2012
Projectile Motion
An object in a projectile motion move horizontally with no acceleration and vertically with the gravitational acceleration at the same time. This experiment is to investigate projectile motion using experiments, equations and comparing the expected and experimental data. Procedure:

Case I:
Use formulas to find equation of horizontal Range (R) in a projectile motion. Rearrange equation for Rmax, and find the angle
Adjust the launches angle to angle
Launch the ball, measure Rmax
Use the equation to solve for initial speed
Case II:
Calculate new R=80/100Rmax
Use to calculate ()
Find out another expected angle , and find its relation with Adjust the launch angle to , launch the ball and measure R
Adjust the launch angle to , launch the ball and measure R
Compare R1 and R2 with R
More Calculation:
Calculate components of velocity for both cases using expected value Calculate maximum height for case I only
Data and Calculations:
In Projectile Motion:
Horizontally (x-direction):
a=0, v=V, X=vt,
Vertically (y-direction):
a=-g, y=vt-1/2gt
Also, v=vcos, v=vsin
We can get R=
Case I:
As -1≤sin2≤1, so sin(2)max=1, so Rmax=v/g, and =45
When the launch angle is 45, Rmax from experiment we get was 1.75m Using equation, we can calculate for V==4.14m/s
Case II:
As, R=, SO expected values of are , and
When the launch angle is 26.6, the range we get R=1.45m;
When the launch angle is 63.4, the range we get R=1.35m.
They are both around the expected range which is R=1.4m
More Calculation:
Components of velocity
For case I:
  For case II:
  When angle is 26.6, v=Vcos26.6=3.7, V=Vsin26.6=1.85 m/s   When angle is 63.4, v=Vcos63.4=1.85, V=Vsin63.4=3.7m/s Maximum height for case I
Rmax=Vt, t= Rmax/V = 1.75/2.93=0.597s
Hmax=V(t/2)-1/2g(t/2)=2.93*(0.597/2)-1/2*9.81*(0.597/2)=0.438m Conclusion:
For case one, we...
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