Yaworski 1
Michael Yaworski
Group Members: Jordan Wickens, Tony Ryu
Mr. Mijic
SPH 3UI
08 Apr 2013
Modelling Projectile Motion
Question:
How did changing the vertical displacement of a projectile affect its range and time of flight?
Hypothesis:
I predict that increasing the vertical displacement of a projectile will increase the projectile's
range and time of flight. The reason for my prediction is that the horizontal displacement (range) of an
object is influenced by its time of flight and and average velocity, as shown in the equation below:
→
→
∆dx = Vavx • ∆t
Because there is no horizontal acceleration (assuming negligible air resistance), the horizontal
velocity will be constant and is also unaffected by vertical displacement. This means that the only
variable that will affect the range of the projectile will be time of flight: the longer the time, the larger
the range; the shorter the time, the smaller the range.
Furthermore, the time of flight is affected by the vertical displacement. Because the only thing
that changes is the increase in the vertical displacement, the projectile will have a greater distance to
fall and will therefore be in the air for a longer period of time. Proof of this is in the equations below:
→ → →
→
∆dy = Viy • ∆t + (½) • a • ∆t2
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In this case, the initial vertical velocity is zero because it is in free-fall. This means that we can
eliminate the term that includes the initial vertical velocity in the equation:
→
→
∆dy = (½) • a • ∆t2
Now to rearrange the equation to solve for time, it would look like this:
→ →
∆t = √[(2• ∆dy ) / g]
Gravity is constant, and therefore, the only variable that will affect the time of flight is the
vertical displacement: the larger the vertical displacement, the longer the time of flight; the smaller the
vertical displacement, the shorter the time of flight.
Materials:
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a ramp
a metal sphere
a desk
three textbooks
a ruler
a metre stick
four sheets of blank paper
a sheet of carbon paper
masking tape
a pencil
Method:
1. On the ramp, a spot was marked with a pencil 30 centimetres from the bottom (while it was
straight).
2. The bottom of the ramp was placed 10cm from the edge of the desk, and the ramp was curved
in such a way that the spot marked on the ramp was perpendicular to the desk and was 20cm
away from the desk.
3. The sphere was rolled from the spot marked on the ramp and a visual estimate of where the
sphere landed was made.
4. Step 3 was repeated three times to get a good estimate of where the ball would land.
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5. A piece of paper was then taped to the floor and the carbon piece of paper was taped over top of
the piece of paper.
6. The sphere was then rolled from the spot marked on the ramp, allowing the sphere to strike the
floor, where it left a carbon dot on the piece of paper.
7. Step 6 was repeated five times (each called a trial) to produce a cluster of carbon dots on the
piece of paper.
8. The horizontal displacements of the projectile was measured and recorded by measuring the
distance from the edge of the desk to each dot of carbon on the piece of paper.
9. The vertical displacement of the projectile was measured and recorded by measuring the
distance from the platform under the ramp to the floor.
10. Step 2 was repeated except with a textbook under the ramp at the very edge of the desk, where
the textbook would act as the desk; step 5 was repeated except replaced with a new piece of
paper; and steps 6-9 were repeated nine more times.
11. Steps 2, 5, 6-9 were repeated four more times except with another textbook placed under the
ramp each time, and a new piece of paper to replace the old one.
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Observations:
Trial Number
Vertical
Theoretical Time
Displacement (m) of Flight (s)
Theoretical Range Experimental
(m)
Range (m)
1
0.757
0.393
0.583
0.585
2
0.757
0.393
0.583
0.576
3
0.757
0.393
0.583
0.585
4
0.757
0.393
0.583
0.587
5
0.757
0.393
0.583
0.583
Initial Velocity (Vix) = 1.484 m/s [down]
Average Range: 0.5832 m
6
0.785
0.400
0.586
0.592
7
0.785
0.400
0.586
0.594
8
0.785
0.400
0.586
0.576
9
0.785
0.400
0.586
0.580
10
0.785
0.400
0.586
0.587
Initial Velocity (Vix) = 1.464 m/s [down]
Average Range: 0.5858 m
11
0.813
0.407
0.594
0.585
12
0.813
0.407
0.594
0.587
13
0.813
0.407
0.594
0.588
14
0.813
0.407
0.594
0.598
15
0.813
0.407
0.594
0.612
Initial Velocity (Vix) = 1.460 m/s [down]
Average Range: 0.594 m
16
0.841
0.414
0.608
0.587
17
0.841
0.414
0.608
0.608
18
0.841
0.414
0.608
0.610
19
0.841
0.414
0.608
0.615
20
0.841
0.414
0.608
0.619
Initial Velocity (Vix) = 1.468 m/s [down]
Average Range: 0.6078 m
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Analysis:
To calculate the theoretical time of flight, the equation discussed in the hypothesis was used:
→ →
∆t = √[(2• ∆dy ) / g ]
Values were then subbed in to solve for the theoretical time of any vertical height. Example:
↓+
∆t = √[(2• 0.757m) / 9.81 m/s² ]
∆t ≈ 0.393s
That method of calculation was done for every change in vertical displacement of the projectile.
As the vertical displacement of the projectile increased, the theoretical time of flight increased
as well. This was because the two variables that affected the time of flight were acceleration and
vertical displacement; and because acceleration was gravity, which was constant, the increase in
vertical displacement correspondingly increased the time of flight. This relates back to the equations
presented in the hypothesis, as well as the equation above, which calculates the theoretical time of
flight.
To calculate the theoretical range of the projectile, a few equations was used. The first one as follows:
→
→
∆dx = Vavx • ∆t
When rearranged, it looked like this:
→ →
∆t = (∆dx / Vix)
The second equation was as follows:
→
→
∆dy = (½) • a • ∆t2
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To solve for the initial horizontal velocity of the projectile at a given height, the first equation was
substituted into the second equation:
→
→ → →
∆dy = (½) • a • (∆dx / Vix)2
The equation was then rearranged to solve for the initial horizontal velocity of the projectile:
→
→ →
→
Vix = √[g • ∆dx2 / (2 • ∆dy )]
The average of the experimental ranges of a given height were calculated, and then subbed into the
equation as the horizontal displacement, as well as gravity and the vertical displacement to solve for the
initial horizontal velocity. Example:
→+
↓+
→
Vix = √[9.81m/s2 • 0.5832m2 / (2 • 0.757s)]
→
Vix ≈ 1.484 m/s
Using that initial horizontal velocity and the theoretical time of flight, the theoretical range was found:
↓+
→
→
∆dx = Vix • ∆t
→
∆dx = 1.464 m/s • 0.393s
→
∆dx = 0.583m
That method of calculation was then applied to every given height.
As the vertical displacement of the projectile increased, the theoretical range increased as well.
This was because the two variables that affected the range of the projectile are the time of flight and
average horizontal velocity. Because the average horizontal velocity was relatively the same for every
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height, even slightly increased, the time of flight more so determined the range of the projectile. The
time of flight was proven to increase as the vertical displacement increased and because the increase in
time of flight increased the theoretical range of the projectile, as the vertical displacement increased,
the theoretical range also increased.
The theoretical range and experimental range were within approximately 1.0 cm of each other;
even more accurate though, the theoretical range for a given was within 0.2 cm of the average of all of
the experimental ranges for that given height. This was because the average range of a given height was
used to calculate the initial horizontal velocity, which was then used to calculate the experimental
range.
Conclusion:
In conclusion, the hypothesis made was correct, in that the increase in the vertical displacement
of the projectile increased the time of flight, as well as the range of the projectile; theoretical and
experimental evidence in the observation table proved this.
The ramp was not held securely in the same position throughout the entire exercise, and so the
difference in the angle of the curve, the impact with the table, the path of the sphere not being
perpendicular to the edge of the desk, etc., led to slightly inaccurate results. More sources of error
would include when the sphere was rolled down the ramp and it scraped itself against the edges of the
ramp to reduce speed. To elaborate on the path of the sphere, if path that the sphere took from the desk
was not exactly perpendicular, the sphere still displaced as usual, but the measurements made were not
accurate because the measurements made were been perpendicular to the edge of the table. This means
that the measurement made was shorter than the actual displacement of the sphere.
If this experiment was to be repeated, the way to decrease the inaccuracy of the experiment
would be to secure the ramp to a fixed position on a textbook, and then simply add a new textbook
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underneath the previous one every time the height is needed to be increased. That would have solved
most of the problems listed above, excluding the sphere that scraped against the ramp. Although it
would not directly solve that source of error, more accurate rolling of the sphere could be done.
If time was to be measured during the experiment, acceleration (gravity) could be calculated. By
using the data already calculated, time could be substituted into the following equation to solve for the
acceleration:
→
→
∆dy = (½) • a • ∆t2
→
→
a = (2 • ∆dy ) / ∆t2