The Acceleration Due to Gravity
Essay Preview: The Acceleration Due to Gravity
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Objective:
In this experiment we will determine that the displacement of a free falling body from the rest is directly related to the square of the falling time and that the acceleration due to gravity does not depend on the mass of the object. We will also determine the experimental value of the acceleration due to gravity.
Theoretical Background:
We can determine the position and velocity of a freely falling object by using the kinematic equations for motion with constant acceleration, since the acceleration due to gravity is constant near the surface of the Earth. As the motion is on the vertical axis, we will designate the displacement by y in the kinematic equations:
v=v_0+gt,
y=v_0 t+〖gt〗^2/2,
v^2=v_0^2+2gy,
where v_0 is initial velocity.
Since we are considering an object that is dropped, the initial velocity v_0 would equal to zero (v_0=0) in the kinematic equations. Therefore, the kinematic equations for the object dropped from rest are
v=gt,
y=〖gt〗^2/2,
v^2=2gy.
Solving the equation for acceleration due to gravity we get
g=2y/t^2 .
Procedure:
Attach a right angle clamp to the vertical support rod while connecting the free fall adapter horizontally in the clamp.
Place the free fall adapters receptor pad on the floor underneath the release mechanism.
Place the metal ball in the release mechanism by tightening the screw on the top of the release mechanism.
Connect the Science Workshop interface box to the computer.
Connect the free fall adapters plug onto Digital Channel 1 on the Science Workshop.
Open Data Studio and click “Create Experiment”.
Double click the “Free Fall Adapter” on the sensor list.
Click on the icon that appears below the Digital Channel 1 of the Interface box 750.
Click “measurement” on the Sensor Property window and check the “Time of fall, Channel 1”.
Click “Start” when ready.
Loosen the screw to release the ball. When the ball hits the receptor pad, click stop. Repeat 3 times for each ball.
Follow step 8 for the heights of 0.5, 0.7, 0.9, 1.0, 1.1, 1.2, 1.3, and 1.4 meters
Sample Calculations:
Height,y=v^2/2g
=〖0.974015〗^2/(2∙9.81)
=0.048354m
Velocity,v=gt
=9.81∙0.099288
=.974015 m⁄s
Questions and Answers:
What kind of information is indicated by the graph of the acceleration due to gravity versus balls mass if the slope of the curve is zero?
If the slope of the curve is zero the acceleration due to gravity is constant.
A ball is dropped from the roof of a building. The ball strikes the ground after 6s.
v_0=0 m⁄s
θ=0 m⁄s
t=6s
v_0y=v_0 sinθ
v_0x=v_0 cosθ
How tall, in meters, is the building?
y=y_0+v_0y t-(gt^2)/2
=0+0(6)-(9.81(6)^2)/2