Acceleration and Free Fall Lab Objective This experiment aims to demonstrate that the free fall acceleration of objects near the earth's surface traveling that move only under the control of earth’s gravitational field is given by the value g = 9.81 m/sec2. Apparatus ο· ο· ο· ο· Digital timer Measuring scale Projectile Calculator Procedure 1. The distance ‘x’ between the first and second photogate was set as 0.20 m and the cylinder was dropped from the release point. 2. Time ‘t’ taken by the cylinder to reach the second photogate was recorded. 3. The acceleration due to gravity ‘g’ was calculated using Equation 1. π= 2π₯ π‘2 (1) 4. Steps 1 and 2 were repeated for four more distance values. 5. The average value of ‘g’ was calculated for the five trials. 6. The percentage error between the accepted and measured value of g was calculated. Results and Discussion The results obtained in this experiment are presented in Table 1. Distance, x (m) 0.20 0.40 0.60 0.80 1.00 Time, t (sec) 0.200 0.287 0.350 0.420 0.450 Average: Acceleration of Gravity, g (m/sec2) 10.0 9.71 9.80 9.07 9.88 9.69 The percentage error between the accepted value of g (9.81 m/s2) and the experimental value is calculated using Equation 2. π΄πππππ‘ππ ππππ’π − ππππ π’πππ ππππ’π πππππππ‘ πΈππππ = | | × 100% π΄πππππ‘ππ ππππ’π π·ππππππ π¬ππππ = π. π% (2) Thus, the percent error between the theoretical value of g and experimental value of g is only 1.2%. Hence, the experimental results agree with the theory and the acceleration due to gravity of an object near earth’s surface is 9.81 m/s2. Question 1. A basketball player jumps straight up to grab a rebound. If she’s in the air for 0.80 seconds, how high did she jump? Solution: Assuming that the player comes down from the peak of her rebound: Initial velocity of player = vo = 0 m/s Time taken by player to come down from peak of her rebound = t = 0.80/2 = 0.40 s Acceleration of the player = a = g = 9.81 m/s2 Using the simple form of 3rd Equation of Motion: β= 1 2 ππ‘ 2 1 (9.81)(0.40)2 2 β= π = π. ππ π Thus, the player jumped 0.78 m high. Question 2: On a manned orbital mission to a planet in the far, far reaches of the galaxy, the brave astronauts observe a geyser spouting from the surface of the planet. Based on measurements the astronauts conclude that the water emerges with an initial speed of 20.00 m/sec and reaches a height of 16.67 meters above the planet. What’s the acceleration of gravity on the surface of the planet? Solution: According to the given information, Initial speed of water = v0 = 20.00 m/sec Height above the planet = x = 16.67 meters Final velocity of water = vf = 0 m/s The acceleration of gravity on the surface of the planet can be calculated using the 4th Equation of Motion π£π2 = π£02 + 2ππ₯ Rearranging the equation for ‘a’, π£π2 − π£02 π= 2π₯ π= 02 − 20.002 2(16.67) π = −12.0 π/π 2 Thus, the acceleration of gravity on the planet’s surface is, π = −π = ππ. π π/ππ Question 3: Are the calculations of the cartoon diver on the right factually accurate according to the laws of physics? How fast will he be going when he hits the water? (Is he really a whiz at physics??? Solution: No, the calculations of the cartoon diver on the right are not accurate according to laws of physics. To calculate his speed upon hitting the water, using 4th Equation of Motion, π£π2 = π£02 + 2ππ₯ π£π2 = 02 + 2(32.2)(50) = 3220 π£π = 56.7 ππ‘/π ππ Converting ft/sec to miles per hour, ππ = ππ. π πππ Thus, he will be going at 38.7 mph when he hits the water. Question 4: In the absence of air resistance, all objects dropped near the Earth’s surface accelerate at the same rate. If two similar objects are dropped, one a short time interval Δt after the other, a. What can you say about the relative speeds of the two objects at any instant after both objects are in free fall? In other words, will ball B be moving faster, slower, or at the same speed as ball A? Why? Answer: The ball that is dropped earlier (Ball B) will be moving faster than the ball that is dropped after a short time interval (Ball A). This is because the speed of Ball B at any time t after both balls are in free fall will be obtained by considering the time t+ Δt while for Ball A, the speed will be obtained by considering only t. As a result, Ball B will have a higher speed at any time t. b. What happens to their separation distance at any instant after both are in free fall? Will it increase, decrease, or stay the same? Why? Answer: Since Ball B is moving with higher speed, the separation between the two objects will increase with time in free fall. c. How does the time interval between the impacts of both objects compare to the time interval between their drop times? Will it greater, less than, or the same? Why? Answer: The time interval between the impact of both objects will be the same as the time interval between their drop times. This is because both objects will cover the same distance and will take the same amount of time to cover that distance. So, their impact time will only differ by the short time interval Δt. Question 5: Something that you’ve all observed: When you turn the water on in your kitchen and slow it down so that the stream flows in a steady, non-turbulent manner, the stream gets narrower as it falls toward the sink. Why does this happen? Answer: This happens because of action of gravity on falling water. As the water falls under the influence of gravity, it accelerates moving down. The particles of water at the bottom have a higher velocity compared to the particles in the upper part of the stream. In order to satisfy the principle of continuity i.e., A1v1 = A2v2, the cross-sectional area of the bottom part of stream has to be smaller than that to the upper part. That is why, the stream gets narrower as it speeds us falling toward the sink.
