Experiment 3: Projectile Motion
The aim of this experiment is to measure height using a ruler and calculate the same height using equations of projectile motion and compare. The equipment needed include
A photo gate, a computer, a stand, a plane wood, a spherical metal ball, vernier caliper, a meter stick, a thin metal rod, a string and a small weight.
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photo gate
- computer
- stand
- plane wood
- spherical metal ball
- vernier caliper
- meter stick
- thin metal rod
- string
- small weight
Theory
A Photo Gate
A photo gate is essentially a stop watch which is triggered by the blocking or unblocking of a light signal that travels from one of its sides to the other side. When in the mode for measuring the speed of small objects that cross it, the stop watch is started when the light signal is blocked and stopped when the light signal is unblocked. Now suppose an object of size d is crossing the photo gate with a speed v. The photo gate will measure the time taken for the size d of the object to cross it. So, if the time measured is t, then the speed may be approximated by
v = d ⁄ t
Even though v is the average velocity over the time interval t, if the object is small, d will be small and v will be a good approximation of the instantaneous velocity of the object as it crosses the photo gate.
Projectile Motion
Projectile motion is motion under gravity. The acceleration due to gravity is directed perpendicularly downward and has a value of -9.8 m ⁄ s 2. Since it doesn’t have a horizontal component, the component of the motion in the horizontal direction is motion with a constant speed while the vertical component of the motion is a uniformly accelerated motion. If an object starts with an initial velocity whose magnitude is vi and makes an angle θi with the positive x-axis, is displaced horizontally by Δx and vertically by Δy; and ends up with a final velocity whose magnitude is vf and makes an angle θf with the positive x-axis in a time interval t, then the motion is governed by the following equations:
vf cos ( θf ) = vi cos ( θi )
Δx = vi cos ( θi ) t
vf sin ( θf ) = vi sin ( θi ) + gt
Δy = vi sin( θi ) + gt 2
{ vf sin ( θf ) } 2 = { vi sin ( θi ) } 2 + 2gΔy
Δy = { vi sin ( θi ) + vf sin ( θf ) } ( t ⁄ 2 )
where the gravitational acceleration g = -9.8 m ⁄ s 2. Now substituting for t from the second equation into the fourth equation of the above equations the following equation can be obtained,
Δy = Δx tan ( θi ) + ( g ⁄ 2) { Δx ⁄ ( vi cos ( θi ) } 2
Therefore, if we know the initial velocity (both magnitude and direction) and the horizontal displacement, we can determine the vertical displacement.
Procedure
To reduce the error statistically, repeat procedures (3.) through (7) five times to fill Table 1.
Table 1
update table