Projectile Motion of launched ball

AI Thread Summary
The discussion focuses on solving a projectile motion problem involving a basketball thrown from a distance of 10 meters to a hoop 14 meters high. The key equations of motion are established, and participants explore how to derive the angle of launch using algebraic manipulation of the equations. A quadratic equation is formed to find the angle, with participants sharing their methods and results from both algebraic and computational approaches. Confusion arises regarding the correct application of gravitational acceleration and the validity of the solutions obtained. Ultimately, the conversation emphasizes the importance of correctly interpreting the equations to arrive at the correct launch angle.
solarwind
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Projectile Motion - Finding The Angle

Homework Statement



You are standing 10 m away from a basketball net that is 14 m above the ground. Assume delta d[y] is 14 m, do not account for your height. You launch the ball with a velocity of 25 m/s and it goes in the hoop.

At what angle(s) could you have thrown the ball so that it goes through?


Homework Equations


The kinematics equations and others that I may not be aware of.
\vec{v}=25 m/s
\Delta\vec{d}[x]=10 m
\Delta\vec{d}[y]=14 m
\vec{a}[y]=-9.8 m/s^2
\theta = ?
 
Last edited:
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Try using the equations of projectile motion:

vx=vCos(theta)
vy=vSin(theta)+0.5at

x=x0+vx0t
y=y0+vy0t+0.5at2
 
Dschumanji said:
Try using the equations of projectile motion:

vx=vCos(theta)
vy=vSin(theta)+0.5at

x=x0+vx0t
y=y0+vy0t+0.5at2


That's exactly what I did but I don't know where to go from there.
 
eliminate t with algebra and then solve for theta.
 
Dschumanji said:
eliminate t with algebra and then solve for theta.

Ok now I have all the equations I need:

V[x]^2 + V[y]^2 = 25^2 = 625
14 = V[y] \Delta t - 4.9 t^2
10 = V[x] \Delta t

I know how to do simultaneous equations. I have to solve for V[x] and V[y] from the second and third equations and plug that into the first one to get time. Then I can plug time into any equation and solve for V[x] and V[y]. Once I have those, the angle is simply Tan-1(V[y]/V[x]).

I did it all in mathematica (math software), but when I tried to actually solve it on paper, I scribbled 6 pages and couldn't solve it. The answer was right when done on the computer but now I have to do it algebraically. Can you help me solve it algebraically?

So,

V[x] = 10/t
V[y] = \frac{4.9t^2 + 14}{t}

Then,

25^2 = V[x]^2 + V[y]^2
25^2 = (10/t)^2 + (\frac{4.9t^2 + 14}{t})^2

This is where it gets messy.
 
What are the answers that mathematica gave you? I would like to see if mine are close after doing it on paper.
 
I put in: Solve[25^2 == (10/t)^2 + ((14 + 4.9 t^2)/(t))^2, t]

It gave me: {{t -> -4.43739}, {t -> -0.791265}, {t -> 0.791265}, {t -> 4.43739}}

These two are the right answers for t: {t -> 0.791265}, {t -> 4.43739}

All I want to know is how to get to this point on paper.
 
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your method of finding theta by first finding the value of t is horribly inefficient.

Start with the equations:

x=x0+vx0t
y=y0+vy0t+0.5at2

We know that at some time t the ball will have the coordinates (10, 14) which translates to:

10=x0+vx0t
14=y0+vy0t+0.5at2

We assume the ball is launched from the origin so the equations simplify to:

10=vx0t
14=vy0t+0.5at2

We are given that the magnitude of the balls velocity when it is launched is 25. That means:

vx0=25Cos(theta)
vy0=25Sin(theta)

The equations now simplify to:

10=25Cos(theta)t
14=25Sin(theta)t+0.5at2

The first equation can be solved for t and then subsituted into the second equation. After some shuffling and applying trig identities you arrive at the quadratic formula:

0=-(14+0.08a)+10Tan(theta)-(0.08a)Tan2(theta)

Substitute u = Tan(theta) into the expression and solve for u. You should get two answers. Use these answers in the equation Tan-1(u)=theta and you should get what you were looking for.
 
your method of finding theta by first finding the value of t is horribly inefficient.
Well sorrrrryyyyyy!

Dschumanji said:
10=25Cos(theta)t
14=25Sin(theta)t+0.5at2

The first equation can be solved for t and then subsituted into the second equation. After some shuffling and applying trig identities you arrive at the quadratic formula:

0=-(14+0.08a)+10Tan(theta)-(0.08a)Tan2(theta)

Substitute u = Tan(theta) into the expression and solve for u. You should get two answers. Use these answers in the equation Tan-1(u)=theta and you should get what you were looking for.
Can you explain how you got from the top two equations to the quadratic equations? I tried it but I'm getting imaginary numbers.
 
  • #10
According to your equation, I did this:
Solve[0 == -(14 + 0.08 (-9.81)) + 10 u - 0.08 (-9.81) u^2, u]

It spits out: {{u -> -13.9493}, {u -> 1.20716}}

So far, I have 3 different sets of answers (6 answers in total) which all seem to work out. There should be only 2 solutions. Which one is right?
 
  • #11
solarwind said:
Well sorrrrryyyyyy!


Can you explain how you got from the top two equations to the quadratic equations? I tried it but I'm getting imaginary numbers.
I didn't mean to be offensive. I just wanted to point out that there was a much easier method.

Solving for t in the first equation gives:
t=0.08Sec(theta)

Inputting this into the second equations gives:

14=25Sin(theta)[0.4Sec(theta)]-0.5a[0.4Sec(theta)]2
14=10Tan(theta)-0.08aSec2(theta)

There is a trig identity that says Sec2(theta) = Tan2(theta)+1 so

14=10Tan(theta)-0.08a[Tan2(theta)+1]
0=-(14+0.08a)+10Tan(theta)-0.08aTan2(theta)

a=9.8
 
  • #12
solarwind said:
According to your equation, I did this:
Solve[0 == -(14 + 0.08 (-9.81)) + 10 u - 0.08 (-9.81) u^2, u]

It spits out: {{u -> -13.9493}, {u -> 1.20716}}

So far, I have 3 different sets of answers (6 answers in total) which all seem to work out. There should be only 2 solutions. Which one is right?
a=9.80 not -9.80
 
  • #13
Dschumanji said:
a=9.80 not -9.80

Thank you! It works now. Why is it not negative?

And thanks again for all your help! It's clear now!
 
  • #14
solarwind said:
Thank you! It works now. Why is it not negative?
My equation already takes into account that the acceleration due to gravity is pointing down, so all that is necessary is putting in the magnitude of the acceleration which is 9.80

Do you have an answer key or something for this problem?
 
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  • #15
Dschumanji said:
My equation already takes into account that the acceleration due to gravity is pointing down, so all that is necessary is putting in the magnitude of the acceleration which is 9.80

Do you have an answer key or something for this problem?

No, but I know your equation is right, I simulated it.
 
  • #16
solarwind said:
No, but I know your equation is right, I simulated it.
It is good to know it works and that I have not been wasting your time with wrong answers!

:approve:
 
  • #17
And where exactly did you modify it so it takes into account that gravity is pointing down?

Edit, nevermind I see it. Thanks so much!
 
  • #18
solarwind said:
And where exactly did you modify it so it takes into account that gravity is pointing down?

Edit, nevermind I see it. Thanks so much!
No problem at all.

night :zzz:
 

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