PHYS 1415V Lab_Projectile_Motion Virtual Lab: Projectile Motion. Questions and worked Solutions.

Virtual Lab: Projectile Motion Name(s): Ana Jimenez, Laura Juarez Date: 12/18/2017 Please use a font color other than black, red or green. Theory: A projectile is an object that moves in two dimensions, that is it m ...

Virtual Lab: Projectile Motion Name(s): Ana Jimenez, Laura Juarez Date: 12/18/2017 Please use a font color other than black, red or green. Theory: A projectile is an object that moves in two dimensions, that is it moves both vertically and horizontally at the same time. A special characteristic of a projectile is that its motion in the vertical direction is independent of its motion in the horizontal direction. If air resistance is small enough to be ignored, then the only force acting on a projectile after it is launched, is the vertical force of gravity. Therefore, in the absence of air resistance, the horizontal component of the projectile’s velocity does not change. In the vertical direction, the force of gravity causes a constant downwards acceleration of 9.8 m/s2. The projectile’s motion in the vertical direction can therefore be analyzed using kinematic equations that are based on constant acceleration such as: y=y0+v0 yt- 12 g t2[ Equation1] where: y = change in height of the projectile (in meters) y0 = initial height of the projectile (before it was launched) voy = initial velocity of the projectile in the vertical direction g = acceleration due to gravity (9.8 m/s2). We put a negative sign in front of g to indicate that we have chosen upwards as the positive direction, and g is directed downwards. t = time in seconds since the launch of the projectile Projectile motion is seen when you throw a ball across to someone, when you hit a baseball or a tennis ball, when water sprays out of a fountain and so on. If a projectile is launched at an angle, we can use trigonometry to find its starting velocities in the horizontal and vertical directions. The figure below explains how. Fig 1. Velocity components and the trajectory of a projectile Fig. 1 shows a red ball being launched with an initial or starting velocity of v0 meters per second at an angle of  measured from the horizontal. This initial velocity can be broken down using trigonometry into its horizontal component (v0cos ) and its vertical component (v0sin). After finding the vertical and horizontal components, different kinematic equations can used to accurately predict the projectile’s path or trajectory. If air resistance is small enough to be ignored (generally the case for a small heavy ball), then the velocity in the horizontal direction does not change and it is calculated using the equation: Horizontal velocity = v0x = v0cos [Equation 2]. Since the horizontal velocity is constant, we can write another useful equation involving velocity, distance, and time: Horizontalvelocity=v0cosθ=Distance time [Equation3] On the other hand, for motion in the vertical direction, we can derive a useful equation by rearranging equation [1] and substituting v0y = v0sin (from Fig. 1) to get: v (¿¿0 sinθ ⁡)t-1 2 g t2[ Equation 4] -h=¿ Here the “-h” is the change in height of the projectile (the height of the floor, which is zero meters, minus the initial height, which is h meters). Let us look at an example to see how these equations come in handy: Example 1: A projectile is fired at an angle of 0 with an initial velocity of 3 m/s. It is launched from a height of 1 meter above the floor. How far will it travel? Solution: We are given the following information:  = 0 v0 = 3 m/s h = 1 m We are asked to find the distance traveled (or the range) in meters. We could use equation [3] to find the range, but we do not know the time the projectile takes to hit the floor. Therefore, we will first have to find the time using equation [4] and then use that time in equation 3. The time the projectile takes to travel the vertical distance to the floor is the same as it takes to travel the horizontal distance before hitting the floor. Equation 4 is: v (¿¿0 sinθ ⁡)t- 1 2 g t2[ Equation 4] -h=¿ substituting the given values and using g = 9.8 m/s2, we get: -1 = (3)(sin0)t – ½(9.8)(t2) The sine of 0 is 0, therefore the whole sin term goes away, leaving us with: -1 = - ½(9.8)(t2) Rearranging this to get t2 on one side we get: t2 = 2/9.8 Taking the square root on both sides we get: t = sq.root(0.2041) t = 0.45 seconds. Now that we have the time, we can substitute it into equation [3] to find the distance: v 0 cosθ= Distance time [Equation3] Substituting values we get: (3)(cos0)= Distance 0.45 The cosine of 0 is 1. Therefore we get: 3= Distance 0.45 which gives us: Distance=(3)(0.45) Distance=1.35 m The projectile will travel a distance of 1.35 meters in the horizontal direction before it hits the floor. We could also say that the range of the projectile is 1.35 meters. Videos: Please watch these short videos for useful information: https://mediaplayer.pearsoncmg.com/assets/free_fall https://mediaplayer.pearsoncmg.com/assets/velocity_projectile Purpose: To investigate how the angle of launch and air resistance affect the motion of a projectile. Lab Procedure: Part 1: Changing the angle, in the absence of air resistance In this experiment, we will see how the angle of launch affects the distance traveled by a projectile. 1. We will use the free simulations provided by the University of Colorado, Boulder. You will need Adobe Flash Player and Java installed on your computer. 2. Open the “Projectile Motion” simulation by Ctrl+ clicking on the link below: https://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html Adjust the simulation settings to the values listed in Table 1 below. Click on “Fire” to run the simulation and record the Range, Height and Time in the last three columns of Table 1. The first two rows have been completed as examples. Table 1 Part 2: Changing the diameter of the ball, in the presence of air resistance and at a constant angle of 45. Please use the animation settings listed in Table 2. Use the same settings as in Table 1 for Projectile type (baseball), initial speed (18 m/s) and mass (0.145 kg). Add “Air Resistance” and change the diameter of the baseball to the different values listed. Run the simulations and record your measurements in the last three columns. The first two rows have been completed as examples. Table 2 Questions: Please select (highlight in a different color) the best answer from the choices provided: 1. From the experiment in Table 1 we can see that the projectile: a) had the largest range when the angle was 90. b) traveled exactly 100 meters when the angle was 20. c) had the smallest range when the angle was 45. d) traveled the farthest from the canon when the angle was 45 2. From the experiment in Table 1 we can see that a) the time the projectile stayed in the air decreased as the angle increased from 0 to 90 b) the time the projectile stayed in the air increased as the angle increased from 0 to 90. c) the time the projectile stayed in the air did not change as the angle increased from 0 to 90. d) the time the projectile stayed in the air increased as the angle was changed from 0 to 45 and then gradually decreased again from 45 to 90. 3. From the experiment in Table 2 we can see that a) as the diameter of the ball increased, the range of the projectile increased sometimes but also decreased sometimes, without any definite pattern. b) as the diameter of the ball increased, the range of the projectile increased. c) as the diameter of the ball increased, the range of the projectile did not change. d) as the diameter of the ball increased, the range of the projectile decreased. 4. From the experiment in Table 2 we can see that a) air resistance does not affect the motion of the ball, the increased diameter of the ball caused it to accelerate in the upward direction. b) air resistance exerts a force on the ball causing it to speed up and travel a longer distance. c) air resistance exerts a force on the ball causing it to slow down and travel a shorter distance d) air resistance does not affect the motion of the ball, the increased diameter of the ball makes it an easier target for gravity to pull down. 5. If we fire an “Adult human” and a “Piano” from the canon, with the same initial speed, at the same angle and in the absence of air resistance, then we can see that (play the simulation to find out and select the correct answer from below): a) the piano does not fit into the canon at all. b) the piano crashes much closer to the canon than the human does. c) the piano lands further away from the canon than the human does. d) the human and the piano have the same range. 6a. Please follow Example 1 in the “Theory” section of this document and solve the following problem. Please show all your steps. Problem: A projectile is fired at an angle of 0 with an initial velocity of 4.2 m/s. It is launched from a height of 1.2 meters above the floor. How far will it travel? 6b. Now check to see if your calculations are correct. Enter the following values into the simulation, click “Fire” and see if your measured range matches your calculated value: Input to simulation: Projectile type: baseball Angle = 0 Initial speed = 4.2 m/s Mass = 0.145 kg Diameter = 0.074 m Air resistance = No The default height is 1.2 meter. 6c. Does your calculated range for the projectile match the measured range? Please enter the names of all group members at the top of this document. Each member of the group is required to submit a copy of the lab report through their eCampus account.