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1. Determine the theoretical range of the launcher for the 3 cases specified bel

ID: 1355080 • Letter: 1

Question

1. Determine the theoretical range of the launcher for the 3 cases specified below.

2. Comment on the accuracy of your results to these theoretical values. What might be causing amy of the differences?

0= y0 + v0sin(theta)T-1/2gT2

Horizontal Velocity: vx=v0cos(theta)

Vertical Velocity: vy=v0sin(theta)-gt

Horizontal Position: x(t)=vxt

Vertical Position:y(t)=y0+v0sin(theta)t-1/2gt2

Yo= 96.7 cm (Distance from floor up to end of barrel)

ax = 0

ay = -g = -9.81 m/s2

Trial 1

30, 2nd notch = 54.3 cm

45, 3rd notch = 68.0 cm

60, 4th notch = 87.1cm

Trial 2

30, 2nd notch = 66.7 cm

45, 3rd notch = 73.2 cm

60, 4th notch = 87.2 cm

Trial 3

30, 2nd notch = 55.7 cm

45, 3rd notch = 79.6 cm

60, 4th notch = 85.1cm

Average Distance

30 = 58.9 cm

45 = 73.6 cm

60 = 86.5 cm

Explanation / Answer

time taken to reach ground, let be t.

vertical position =0.967+v0*sin(theta)*t-4.905*t^2

hence 0=0.967+v0*sin(theta)*t-4.905*t^2


==>4.905*t^2-v0*sin(theta)*t=0.967....(1)


horizontal distance=v0*cos(theta)*t

let horizontal distance be denoted by D.

then t=D/(v0*cos(theta))

substituting in equation 1,


4.905*(D^2/v0^2)*sec^2(theta)-v0*tan(theta)*(D/v0)=0.967

==>4.905*(D^2/v0^2)*sec^2(theta)=0.967+D*tan(theta)

==>v0^2=(4.905*D^2*sec^2(theta))/(0.967+D*tan(theta))

so for each trial, we will substitute value of D and theta to find corresponding value v0 and compare with the value obtained with average distance.


for theta=30 degree:

average distance=D=0.589 m
==>v0=1.3175 m/s


trial 1:
D=0.543 m
==>v0=1.2272 m/s


trial 2:

D=0.667 m
v0=1.4669 m/s
trial 3:

D=0.557 m
v0=1.2548 m/s

for theta =45 degrees:

average value:

D=0.736 m

v0=1.7665 m/s

trial 1:

D=0.68 m

v0=1.6596 m/s


trial 2:

D=0.732 m

v0=1.7589 m/s

trial 3:

0.796 m
v0=1.8777 m/s


for theta=60 degrees:

average value:

D=0.865 m

v0=2.44 m/s


trial 1:

D=0.871 m

v0=2.452 m/s

trial 2:

D+0.872 m

v0=2.454 m/s


trial 3:

D=0.851 m


v0=2.4127 m/s