In this manner e concluded that for an object undergoing projectile motion, the horizontal position changes at a constant linear rate while the vertical position changes exponentially in a square-like fashion. Introduction A toy company wants to produce an instructional video for jugglers to learn about projectile motion.
Because juggling involves a ball flying through the air, knowledge of how a projectile’s horizontal and vertical velocity components behave is beneficial for predicting the position of the ball as it moves through air.In this experiment, a ball was tossed in the air so as to simulate projectile motion notations. Using video software, the ball’s flight was analyzed to determine how the object’s velocity and position components change. Prediction The velocity of an object as a function of time is expressed in equation 1.
1) Equation 1: Velocity of an object as a function of time. If, is final velocity, Vow, is initial velocity, a, is acceleration, t, is time Using trigonometry, we resolved velocity into its components giving us equations 2 and 3. ) Equation 2: Trigonometry was used to determine the horizontal component of velocity. Vs.., is initial horizontal velocity, V, is initial velocity, and , d, is the launch Engle from the horizontal Equation 3: Trigonometry was used to determine the vertical component of velocity. Ivy, is initial vertical velocity, V, is initial velocity, and , B, is the launch Substituting our calculated equations of Vs..
And Ivy from equations 2 and 3 for Vow in equation 1 allowed us to compute equations 4 and 5 representing the vertical and horizontal velocity components as a function of time.Equation 4: Horizontal velocity as a function of time. Vs.
., is final horizontal velocity, Viscose. Is initial horizontal velocity, a, is acceleration and ,t, is time. Equation 5: Vertical velocity as a function of time.
Vhf, is final vertical velocity, Vision. Is initial vertical velocity, a, is acceleration and is time. In the x direction, we predicted that there was no acceleration giving equation 6 from equation 4 6) Equation 6: Horizontal velocity in the absence of horizontal acceleration. Vs.., is the final horizontal velocity , Viscose, is the initial horizontal velocity.Thus, we predicted that the horizontal velocity component of an object undergoing projectile motion will be constant and equal to its initial velocity. This is represented by Graph 1.
Graph 1: Horizontal velocity as a function of time. In the absence of horizontal acceleration, an object’s velocity stays constant and is equal to Voss. V, is initial velocity, B, is the launch angle from the horizontal. Conversely, for the vertical direction, we predicted that gravitational acceleration affects the vertical velocity of the ball. Equation 4 was calculated from equation 7 and it represents the ball’s vertical velocity component.
) Equation 7: Vertical velocity in the presence of a vertical gravitational acceleration. Vhf, is the final vertical velocity, Vision, is the initial vertical velocity , , is gravitational acceleration is time. We determined that during the first half of the projectile’s flight the velocity will decrease at a constant rate and during the second half of the projectile’s flight, the velocity will increase at the same constant rate because of the constant acceleration magnitude. Using this knowledge and equation 7, we sketched graph 2 representing vertical velocity as a function of time.
We determined that because acceleration is constant in the vertical direction, the change in our velocity, or acceleration, will be constant giving us a graph with a constant slope. Graph 2: Vertical velocity as a function of time. Vhf, is final vertical velocity, Vision, is initial vertical velocity, a, is acceleration due to gravity and is 9. Mm/so, , t, is time. Using our knowledge of kinematics, we determined the following equations representing the horizontal and the vertical position components of a projectile as a function of time. 8) Equation 8: Horizontal position as a function of time.X, is final horizontal position, ox, is initial horizontal position, viscose, is initial horizontal velocity and ,a, is acceleration, t, is time.
Again, we predicted that there is no acceleration n the horizontal direction so we simplified equation 8 to equation 9 describing the object’s horizontal position as a function of time. 9) Equation 9: Horizontal position as a function of time in the absence of horizontal acceleration. X, is final horizontal position, ox, is initial horizontal position, viscose, is initial horizontal velocity and,t, is time.From this equation, we were able to draw graph 3. Graph 3: Horizontal position as a function of time for an object in projectile motion. Conversely, we can expressed the vertical position of the projectile as equation 10. Equation 10: Vertical position as a function of time. , is final vertical position, you, is initial vertical position, vision, is initial vertical velocity and,a, is acceleration due to gravity and is equal to 9.
Mm/so, t, is time. Equation 10 was then graphed to predict how the vertical position of a projectile changed as a function of time.Due to the presence of acceleration in the vertical position, we figured that it’s velocity would change. We hypothesized that the direction of the vertical velocity vector is upwards during the first half of a projectile’s flight; its motion is in the upward direction. However, during the second half of a projectile’s flight, the erection of the vertical velocity vector is downward because of its downward motion. In contrast, the direction of the gravitational acceleration vector always points downwards and has a constant magnitude.So, in the first half of the ball’s flight, the object’s velocity would decrease to zero while it would increase from zero in the second half of the ball’s flight. The manner in which velocity changes, first decreasing and then increasing, due to acceleration in the vertical direction prompted us predict a parabola-shaped graph for vertical position vs.
. Time as can be seen in graph 4. Graph 4. Vertical position as a function of time for an object in projectile. This is represented by . Y, is final vertical position, you, is initial vertical position, vision, is initial vertical velocity and,a, is acceleration due to gravity and is equal to 9. M/ so, t, is time. Procedure: A ball was then thrown and its trajectory recorded with a video camera.
Two meter sticks were used for calibration purposes. One was placed in a horizontal orientation, while the second placed at the CACM mark of the first in a vertical orientation. Several practice throws were carried out in order to attempt to get he trajectory of the ball to start at the Com mark of the horizontal meter sticks.
In this manner, the balls highest point could coincide with the plane of the vertical meter stick allowing for accurate video analysis of the object.The ball was thrown and its motion recorded until this was achieved. The ball’s motion was then analyzed with Emotional and used to generate motion graphs ( horizontal displacement vs.. Time, y displacement vs.
. Time, vs.. vs.. Time, and ivy vs.
. Time). These motion graphs were best-fit to compare our predicted values from our experimental values. Data: Figure 1: Top left: horizontal component of the ball’s trajectory vs.. Time graph. Top right: Vertical-component of the ball’s trajectory vs.
. Time graph. Bottom left: Horizontal velocity of the ball’s trajectory vs.
. Time graph.Bottom right: Vertical velocity of the ball’s trajectory vs.. Time graph. Analysis Our predicted equation for the horizontal position corresponding to equation 9 was: . Jazz From our measurements, the value of -1.
Mm/s comes from our calculated Voss. Our experimental data, however, gave us a best-fit equation of: F(z) =. 299-1. Oz Thus we had an error in calculation as our true initial horizontal velocity, or Voss, was -1. Mm/s, while our initial horizontal position was . 299 meters away from where we marked our launch origin in Emotional.For our vertical position, our prediction corresponding to equation 10 was: F(z)=3.
Oz +-4. Oz Because we predicted the presence of gravitational acceleration in the vertical direction , a , or 9. Mm/so; we calculated -4. Mm/so as the value of 9. 8/2.
Obviously this was a calculation error as -9. 8/2 is equal to-4. 9 as seen by our best-fit equation y: Additionally, we slightly miscalculated our initial vertical velocity as evidenced by the difference from our predicted velocity, or 3.
4 m/s, to our actual velocity, or 3. 88 m/s.In hindsight, the calculation errors which made our predicted equations vary from our best-fit equations where caused by inaccuracy in determining the launch angle and velocity of the object. We calculated both by measuring the distance the projectile covered in the x and y directions directly from our computer monitor and divided it by the time given from our video recording. This method was not ideal as our software quality prevented us from being as accurate as possible. Our experimental position and velocity graphs where similar to the graphs we sketched in our predictions.Using calculus, we came up with an explanation for the shapes of our graphs.
We determined that the derivative of position, r, gives us instantaneous velocity,v, as shown by: Likewise, the derivative of instantaneous velocity gives us instantaneous acceleration depicted by: . This knowledge allowed us to analyze our graphs accurately The experimental horizontal position vs.. Time graphs had a negative constant slope. Based on calculus, we know that taking the derivative of position gives us the slope of the line tangent to that point, or instantaneous velocity, at the desired time.Thus, the constant slope of our horizontal position vs.. Time graph meant that our horizontal velocity was constant.
Looking at our experimental horizontal velocity graph we know this is correct as Vs.. Stays at -1. Mm/s as time increases. We can conclude that we were correct in our predictions that ax, or the horizontal acceleration of a projectile, is non-existent. To reiterate, no acceleration in the horizontal direction meant that horizontal velocity of the object stayed constant and so the slope of a position vs.. Time graph didn’t change In the vertical position, our graph was an upside-down parabola; just as we redirected.
If were to draw lines tangent to the vertical position vs.. Time curve, we would see their slope become flatter as time increases until it becomes perfectly horizontal at the peak of the projectile’s motion at ”. 4 seconds.
From . Seconds to . Seconds, the opposite is true; the slope of line would start horizontally and become steeper. From calculus, we know that these tangent lines to vertical position correspond to the behavior of vertical velocity. During the first half, or from Seconds to . Seconds, vertical velocity decreases.
This is because during his time interval the direction of the vertical velocity is opposite to the direction of gravitational acceleration. In a sense, acceleration is decreasing the object’s velocity until it falls to O m/s. This can be seen from our vertical velocity vs.
. Time graph; at . Seconds vertical velocity is mm/s. During the second half of the flight the opposite is true. Velocity starts from rest and increase because the orientation of vertical velocity and acceleration are in the same direction as the projectile plummets to earth.This is evident as the magnitude of the velocity increases to -4.
Mm/s at . Seconds. It is important to note that the slope of vertical velocity is non-zero and constant. Again, from calculus we know that this is due to the presence of a constant gravitational acceleration which changes the projectile’s vertical velocity giving a parabola-like graph of vertical position vs.. Conclusion This experiment has given us the opportunity to explore the movement of a ball undergoing projectile motion as we saw how the vertical and horizontal components of the ball act independently of each other.
Based on our experimentally generated graphs, we justified our predictions and discovered hat in the vertical direction, gravitational acceleration causes deceleration in the first half of the projectile’s flight and acceleration in the second half of the projectile’s flight. Thus, the velocity falls to zero in the first half of the trajectory as its position increases less and less until the object’s peak.