In this lab, you will perform graphical and mathematical analysis on a tennis ball that is struck by a bat.
The data required for this lab is the horizontal displacement of the ball from the location where it was hit to the point where it first strikes the ground and the time that the ball spends in the air. You may assume that the ball was hit from an initial height of 1.5m
t = 3.60 s
△X = 48.9 m
Analysis:
Mathermatical:
*Determine the initial velocity of the ball as it left the bat.
*Determine the maximum height reached by the ball.
*Determine the horizontal distance at which the maximum height was achieved.
*Divide the time that the ball spent in the air into 10 equal interval.
*Determine the displacement, velocity, and acceleration of the ball in both the X and Y directions for each of these time intervals.
*Display the information above in a table.
*Show one set of sample calculations.
Graphical:
*Contrast (x vs t),(v vs t), and (a vs t) graphs for the ball in both the X and Y directions.
*Describe the shape of each graph and explain why each graph has the shape that it does.
Conclusion:
Write summary of your findings from this exercise.
Please Help, answer.
Thank You
Dx = Xo * T = 48.9 m.
Xo * 3.60 = 48.9
Xo = 13.58 m/s = Hor. component of initial velocity.
Tr = Tf = T/2 = 3.60/2 = 1.80 s. = Rise
and fall times.
Y = Yo + g*Tr = 0 @ hmax.
Yo - 9.8*1.8 = 0
Yo = 17.64 m/s = Ver. component of initial velocity.
tan A = Yo/Xo = 17.64/13.58 = 1.29897
A = 52.4o
*Vo = Yo/sin A = 17.64/sin52.4 =
22.3m/s[52.4o]
h max = (Y^2-Yo^2)/2g
*h max = (0-17.64^2)/-19.6 = 15.88 m.
*Dx = Xo * Tr = 13.58m/s * 1.8s=24.44 m.
Use the following time intervals for graphing:
T/10 = 3.60/10 = 0.36 s.
*0.36,0.72,1.08,1.44,1.80,2.16,2.52,2.88,3.24,3.60 s.
To determine the initial velocity of the ball as it left the bat, you can use the equation for horizontal displacement: ΔX = Vx * t, where ΔX is the horizontal displacement, Vx is the initial velocity in the x-direction, and t is the time. Rearranging the equation, you have Vx = ΔX / t. Plugging in the given values, you get Vx = 48.9 m / 3.60 s.
To determine the maximum height reached by the ball, you can use the equation for vertical displacement: ΔY = Vyo * t + (1/2) * a * t^2, where ΔY is the vertical displacement, Vyo is the initial vertical velocity, t is the time, and a is the acceleration due to gravity (-9.8 m/s^2). Since the ball was hit from an initial height of 1.5 m, you can calculate the initial vertical velocity, Vyo, as follows: Vyo = (-1.5 m) / t + (1/2) * a * t. Plugging in the given value of t, you can find Vyo.
The horizontal distance at which the maximum height was achieved is equal to the horizontal displacement when the vertical velocity is zero. You can use the equation ΔX = Vx * t to calculate ΔX when Vyo = 0.
To divide the time that the ball spent in the air into 10 equal intervals, you can divide the total time (3.60 s) by 10. This will give you the length of each time interval.
To determine the displacement, velocity, and acceleration of the ball in both the x and y directions for each of these time intervals, you can use the kinematic equations. The equations for displacement, velocity, and acceleration in the x-direction are the same as in the y-direction, since there is no external force acting horizontally. For each time interval, you can use the equations: ΔX = Vx * t + (1/2) * ax * t^2, Vx = Vxo + ax * t, and ax = 0 (since there is no external force acting horizontally). Similarly, for the y-direction, you can use the equations: ΔY = Vyo * t + (1/2) * a * t^2, Vyo = Vyo + a * t, and a = -9.8 m/s^2 (due to gravity).
To display the information above in a table, you can create a table with columns for time, displacement in the x-direction, displacement in the y-direction, velocity in the x-direction, velocity in the y-direction, acceleration in the x-direction, and acceleration in the y-direction. Then, you can fill in the values for each time interval using the equations mentioned earlier.
For the graphical analysis, you can plot the following graphs: x vs t, v vs t, and a vs t for both the x and y directions.
In the x vs t graph, you plot the horizontal displacement (ΔX) of the ball on the y-axis against the time (t) on the x-axis.
In the v vs t graph, you plot the velocity (Vx) of the ball in the x-direction on the y-axis against the time (t) on the x-axis.
In the a vs t graph, you plot the acceleration (ax) of the ball in the x-direction on the y-axis against the time (t) on the x-axis.
Similarly, you can plot the x vs t, v vs t, and a vs t graphs for the y-direction using the corresponding variables.
Describing the shape of each graph and explaining why each graph has the shape that it does will depend on the specific values calculated. For example, in the x vs t graph, you might observe a linear relationship if the ball moves at a constant velocity in the x-direction. In the v vs t graph, you might observe a constant value if the horizontal velocity remains constant. In the a vs t graph, you might observe a horizontal line at zero if there is no horizontal acceleration.
Finally, in the conclusion, you can summarize the findings from the exercise, including the initial velocity of the ball, maximum height reached, the horizontal distance at which the maximum height was achieved, and any observations from the graphical analysis of the x and y directions.