23-Mar-2015: Predicting impact points of trajectories.

     The purpose of this experiment was to use the understanding of projectile motion to predict the impact of a ball on an inclined board. The setup for this was quite simple. Two ramps were used, one lying flat and the other attached at an inclined angle so a ball could roll down. The flat ramp was set up flush with the edge of a table.

setup of ramps

     Once the ramps were setup, a steel ball was released from the top of the inclined ramp. It rolled down and flew off. The point at where the ball hit the floor was noticed, and regular printing paper, with a piece of carbon paper over it, were put on that spot. This was done so when the ball hits the carbon paper it makes a mark on the regular printing paper. The ball was launched five times to ensure it hits roughly the same spot each time.

     The measurements of the height of the ramp and the distance from it where the ball lands were needed for future calculations. A meter stick was used to measure both. The height of the ramp was measured to be 0.874 meters and the distance from the ramp where the ball hit the floor was 0.734 meters. It might be hard to see in the picture (below and to the right), but there are a small cluster of dark spots where the ball hit the paper.

Measurement of the height of the ramp from the floor.

Measurement of the distance from the ramp where the ball hits the floor.

  

     The next step was to calculate the launch speed of the ball using the measurements taken above. The equation delta x=V0x*t was used since the ramp it was launching off of was flat. First, the time it was in the air needed to be calculated. This was done by using the formula

delta y=V0y*t+(1/2)gt 

and assuming constant acceleration in the y-direction. Since the initial velocity in the y-direction is zero, the formula can be rearranged and solved for t, so that

t=[(2*delta y)/g]^(1/2)

Delta y was a measured quantity, so after plugging in the numbers, the time was .422 seconds. That number was plugged into the formula delta x=V0x*t, which was solved for V0x so that

V0x=(delta x)/t

Delta x was a measured quantity and t was solved for in the previous step. Plugging in the numbers gave an initial x-velocity of 1.74 m/s.

     The next step was to place a wood board at the edge of the table and incline it so that when the ball is launched, the ball will hit the wood board instead of the ground. Before starting the trial, a prediction was made as to where the ball would hit the wood board. This prediction was done by using the measured quantities from earlier and using trigonometry. Below is the picture of the setup of the inclined board.      

     As you can see, the bottom of the inclined board was positioned where the ball would strike the ground. This allowed for the finding of the angle between the board and the ground using 

tan(theta)=y/x

in which y and x were the measured quantities from earlier. The angle was calculated to be 46.3 degrees. Finding the angle then allowed for the prediction of the distance the ball would hit on the board. First, note that y=dsin(theta) and x=dcos(theta). The kinematic equations 

y=(1/2)gt^2

x=V0t 

were used to make the prediction. Substituting dsin(theta) and dcos(theta) for y and x, respectively, gives new equations of 

dsin(theta)=(1/2)gt^2

dcos(theta)=V0t 

The equations were solved for t^2 and set equal to each other, so that

(2dsin(theta))/g=(d^2cos^2(theta))/V0^2

d was then solved for giving a final equation of 

d=(2V0^2sin(theta))/(gcos(theta))

     After plugging in the numbers, the ball was predicted to hit the board 0.936 meters down. 

     After the prediction was made, it was time to start the trial. The piece of printing paper, with carbon paper over it, was positioned around the predicted impact point. The ball was then launched five times from the same spot to make sure it hits roughly the same point on the board. After all trials were completed, the point where the ball hit the board was measured using a meter stick. The distance was recorded to be .890 meters. 

     The final numbers of a predicted measurement of 0.936 m and an actual measurement of 0.890 were not terribly close. But, remembering that there were measurements made, and therefore uncertainties in those measurements, gave some hope that the predicted measurement was within some range of the uncertainty. The total uncertainty was calculated using the usual propagated uncertainty formula, where the partial derivative is taken with respect to some variable, and that partial derivative is multiplied by the uncertainty in the measurement of the variable. The distance that was solved for earlier, however, only had one variable in it, theta. So, some algebraic manipulation was needed to get the x and y variables in the formula for distance. Using 

y=(1/2)gt^2

t was solved for so that t=(2y/g)^(1/2). Then, using 

x=V0t

V0 was solved for so that V0=x/t. At that point, t was plugged into this equation to give 

V0=x/(2y/g)^(1/2)

This V0 was plugged into the equation 

 d=(2V0^2sin(theta))/(gcos(theta))

and simplified to a final equation of 

d=(x^2/y)sec(theta)tan(theta)

     Now that the three measurements with uncertainty are in the formula for distance, the partial derivatives were taken. All numbers were plugged in to solve for propagated uncertainty, which was (plus or minus) 0.0477 m. It worked! The propagated uncertainty gave a predicted range of 0.883 m to 0.9837 m, in which the actual impact measurement was within. There were ways in which this experiment could have had greater errors, such as, not releasing the ball from the same height each time or errors in reading measurements. There were also assumptions made, such as no air resistance when the ball was in its trajectory, and no friction when the ball was rolling down the ramps.