"How to find the initial velocity Vo and the angle of projection of a projectile?"
Materials needed for the experiment:
- A Basketball
- A camera
-A program which helps in modeling the trajectory using the program "Atelier Scientifique".
Procedure:
Take a video of a basketball shot and use a program to make a model of the trajectory of the basketball.
The model of the trajectory should be a parabola.
Start finding the the vector V of the initial velocity. For this, we plot two different graphs : one to find the X coordinate Vx and the other to find the Y coordinate Vy. To find these coordinates, we calculate the derivative of 0 in each graph.
1st Graph: X axis against time (t) to find the X coordinate of v: Vx
The graph showed a straight line with the equation:
f(x)= 4.534x - 0.033
f'(x)= 4.534
f(0)= 4.534
Thus, the X coordinate of v, Vx= 4.534
2ng Graph: Y axis against time(t) to find the Y coordinate of v: Vy
The graph showed a parabola with the equation:
g(x)= -4.615t^2 + 5.755t + 0.023
g'(x)= -9.23t + 5.755
g(0)= 5.755
thus v:
From this, we can calculate the angle of projection Ѳ and the initial velocity Vo of the projectile.
We plot the Y-axis against the X-axis.
tan Ѳ= opp/adj
tan Ѳ= Vx/Vy = 5.755/4.534 = 1.269
tan ^-1(1.269)=51.7°
By placing the protractor on the graph we could say that the angle of projection was 51 ° There was thus a slight deviation in our calculation than the reality (which is quite normal). Thus
The initial velocity Vo is the length of the vector v. Using the Pythagoras theorm, we write:
For the parabola modeling the graph of the Y axis plotted against the X axis in the video, the equation was in the form:
h(x)= ax^2 +bx +c with a= -0.233
According to the formula that we previously calculated, a is also equal to -g/2(Vo*cosѲ)^2
We then substitute the values of Vo and Ѳ to verify the value of g (9.8m.s^-2) and thus to verify the credibility of our calculations.
-0.233 = -g/2(Vo*cosѲ)^2
g= 0.233*(7.3)^2*(cos(51))^2
g= 9.83 m.s^-1/2 , a value quite close to the real constant.
Materials needed for the experiment:
- A Basketball
- A camera
-A program which helps in modeling the trajectory using the program "Atelier Scientifique".
Procedure:
Take a video of a basketball shot and use a program to make a model of the trajectory of the basketball.
The model of the trajectory should be a parabola.
Start finding the the vector V of the initial velocity. For this, we plot two different graphs : one to find the X coordinate Vx and the other to find the Y coordinate Vy. To find these coordinates, we calculate the derivative of 0 in each graph.
1st Graph: X axis against time (t) to find the X coordinate of v: Vx
f(x)= 4.534x - 0.033
f'(x)= 4.534
f(0)= 4.534
Thus, the X coordinate of v, Vx= 4.534
2ng Graph: Y axis against time(t) to find the Y coordinate of v: Vy
g(x)= -4.615t^2 + 5.755t + 0.023
g'(x)= -9.23t + 5.755
g(0)= 5.755
thus v:
From this, we can calculate the angle of projection Ѳ and the initial velocity Vo of the projectile.
We plot the Y-axis against the X-axis.
tan Ѳ= opp/adjtan Ѳ= Vx/Vy = 5.755/4.534 = 1.269
tan ^-1(1.269)=51.7°
By placing the protractor on the graph we could say that the angle of projection was 51 ° There was thus a slight deviation in our calculation than the reality (which is quite normal). Thus
The initial velocity Vo is the length of the vector v. Using the Pythagoras theorm, we write:
h(x)= ax^2 +bx +c with a= -0.233
According to the formula that we previously calculated, a is also equal to -g/2(Vo*cosѲ)^2
We then substitute the values of Vo and Ѳ to verify the value of g (9.8m.s^-2) and thus to verify the credibility of our calculations.
-0.233 = -g/2(Vo*cosѲ)^2
g= 0.233*(7.3)^2*(cos(51))^2
g= 9.83 m.s^-1/2 , a value quite close to the real constant.
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