Projectile motion- initial velocity and launch angle


by mmoadi
Tags: angle, initial, launch, motion, projectile, velocity
mmoadi
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#1
Oct14-09, 10:05 AM
P: 157
1. The problem statement, all variables and given/known data

Basketball player throws the ball form the middle of the court one second before the game ends. How much must the initial velocity of the ball be and at what angle it must be thrown, if it should hit the basket at the exact time of siren without preceding reflection from board? Length of playground is 26 m, the distance between the floor and the ring is 306 cm, and the ball is thrown from height 206 cm. Neglect the rotation of the ball. At what angle and with what velocity the ball hits the basket?

2. Relevant equations

Δx= v0xt
Δy= v0yt (gt)
v0= sqrt(v0x + v0y)
tan α= v0y / v0x


3. The attempt at a solution

What I know:
t= 1 s
L= 26 m
Δx= 13 m
h1= 206 cm
h2= 306 cm
Δy= 100 cm = 1 m

What Im looking for:
v0= ? m/s (initial velocity)
α = ? (launch angle)
vf= ? m/s (final velocity)
β= ? (impact angle)

① First, I calculated the v0x = vx: v0x= Δx / t = 13 m/s
② Second, I calculated v0y: v0y= (2Δy + gt) / 2t= 5.4 m/s
③ I know calculated v0 from the Pythagoras formula: v0= sqrt(v0x + v0y)= 14.08 m/s
④ And know I calculated the launching angle α: α= (tan)-(v0y / v0x)= 22.56

What is left know are final velocity and impact angle. Can I just state that final velocity = initial velocity (conservation of energy) and that impact angle = launching angle (symmetry of parabola)? Or, do I have to do new calculations because the basketball ring is higher than the launching height of the ball and the parabola is not symmetrical? If second is the case, I really need help, because I dont know hoe to approach it now.

Thank you for helping!
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rl.bhat
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#2
Oct14-09, 10:40 AM
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Vy(t) = Vo(y) - gt.
Using the above equation find vy. Vo(x) is known. Find v and angle.
Andrew Mason
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#3
Oct14-09, 10:44 AM
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To solve this problem you need an expression for height as a function of time. You end up with a quadratic equation in terms of v0y and t. Solve that for v0y when h = 1 m and t=1 sec.

AM

mmoadi
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#4
Oct14-09, 01:03 PM
P: 157

Projectile motion- initial velocity and launch angle


So, my calculations are not correct?

I used this formula: Δy= v0yt – (gt)
mmoadi
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#5
Oct14-09, 01:07 PM
P: 157
Quote Quote by mmoadi View Post
Δy= v0yt (gt)

② Second, I calculated v0y: v0y= (2Δy + gt) / 2t= 5.4 m/s
I used the above formula and calculated v0y. Are my calculations correct?
Andrew Mason
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#6
Oct14-09, 01:07 PM
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Quote Quote by mmoadi View Post
So, my calculations are not correct?

I used this formula: Δy= v0yt (gt)
This is correct. Just plug in values for Δy (1 m.) and t (= 1 sec.) to get the value for v0y. You can determine the launch angle from that and v0x (13 m/sec).

AM
Andrew Mason
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#7
Oct14-09, 01:14 PM
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Quote Quote by mmoadi View Post
I used the above formula and calculated v0y. Are my calculations correct?
The formula is correct. But the answer is 11.8/2 = 5.9 m/sec.

It is simpler to write it as v0y = (1 + .5gt^2)/t.

AM
mmoadi
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#8
Oct14-09, 01:29 PM
P: 157
So, now that I have the v0x (13 m/s) and v0y (5.9 m/s) I can calculate the initial velocity v0 form Pythagoras formula v0= sqrt(v0x + v0y) and the launch angle from formula
tanα= v0y / v0x?
Andrew Mason
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#9
Oct14-09, 02:24 PM
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Quote Quote by mmoadi View Post
So, now that I have the v0x (13 m/s) and v0y (5.9 m/s) I can calculate the initial velocity v0 form Pythagoras formula v0= sqrt(v0x + v0y) and the launch angle from formula
tanα= v0y / v0x?
Correct. Can you determine vy and angle at t = 1 sec?

AM
mmoadi
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#10
Oct14-09, 02:36 PM
P: 157
vy= v0y - gt and for t I'm using 1/2 s, which is the time needed for the ball to rise to the max height.
So, vy= 5.9 m/s - (9.8 m/s * 1/2 s)= 1 m/s

And, if I plug this into the Pythagoras formula, the initial velocity should be 13.02 m/s?

As for the angle, is this correct: α= (tan)-(v0y / v0x)= 24.49 ?
Andrew Mason
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#11
Oct14-09, 02:51 PM
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Quote Quote by mmoadi View Post
vy= v0y - gt and for t I'm using 1/2 s, which is the time needed for the ball to rise to the max height.
First of all, why are you trying to find the time needed for the ball to reach maximum height? Second, this is not the way to find it. You have to work out what t is when vy=0. You know that vy = 0 when maximum height is reached. So t = v0y/g = 5.9/9.8 = .60 sec. But, as I say, there is no real need to find this value.

To find vy at t = 1 sec. use your formula for vy. That will enable you to find the angle.

AM
mmoadi
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#12
Oct14-09, 02:58 PM
P: 157
I'm really confused now.
vy= 5.9 m/s - (9.8 m/s * 1s)= - 3.9 m/s.

And for the angle: α= (tan)-(vy / vx)= 16.7
Andrew Mason
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#13
Oct14-09, 03:23 PM
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Quote Quote by mmoadi View Post
I'm really confused now.
vy= 5.9 m/s - (9.8 m/s * 1s)= - 3.9 m/s.

And for the angle do I use the formula sinα= gΔx / v0?

So what do I use for v0?
It is not that difficult if you stick to the correct equations.

[tex]v_y = v_{0y} - gt[/tex]

gives you the vertical component of the velocity of the ball at time t. You have used this correctly to find the vertical speed at t=1 seconds.

You have already found v0 (the launch speed). What you have to do is find the speed at the basket, lets call that vb. You know vby and vbx so calculate vb.

How is the angle at the basket related to vby and vbx?

AM
mmoadi
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#14
Oct14-09, 07:21 PM
P: 157
So it continues:
"You have already found v0 (the launch speed)." Does this mean that the v0 is -3.9 m/s?

vb= sqrt(vbx + vby)= sqrt(169 + 15.21)= 13.57 m/s
Is vb the final velocity of the ball?

As for the angle, I think that the relationship is tan(angle)= vby / vbx, which gives us an angle of -16.7?
Andrew Mason
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#15
Oct14-09, 07:45 PM
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Quote Quote by mmoadi View Post
So it continues:
"You have already found v0 (the launch speed)." Does this mean that the v0 is -3.9 m/s?
NO!! 3.9 m/sec is the vertical component of the speed at the basket. v0 is the launch speed (the vector sum of v0y and v0x).

vb= sqrt(vbx + vby)= sqrt(169 + 15.21)= 13.57 m/s
Is vb the final velocity of the ball?
Yes.
I am not sure why this is not obvious. You know vbx is 13 m/sec at all times. vby is simply vy at time = 1 sec. which you have determined is 3.9 m/sec. vb is simply the vector sum of these orthogonal components. You will notice a lack of symmetry here. The peak of the trajectory does not occur at the midpoint between launch and basket. So the speed at the basket is not the same as the speed at launch. You have to stick to the equations of motion.
As for the angle, I think that the relationship is tan(angle)= vby / vbx, which gives us an angle of -16.7?
Correct.

AM
mmoadi
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#16
Oct15-09, 09:52 AM
P: 157
Quote Quote by Andrew Mason View Post
NO!! 3.9 m/sec is the vertical component of the speed at the basket. v0 is the launch speed (the vector sum of v0y and v0x).
I hope I understood correctly.

v0= sqrt(v0x + v0y)= sqrt(169 + 34.81)= 14.28 m/s

As for the launch angle is this calculation correct: α= (tan)-(v0y / v0x)= 24.49 ?

Please. let me be correct!
Andrew Mason
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#17
Oct15-09, 10:52 AM
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Quote Quote by mmoadi View Post
I hope I understood correctly.

v0= sqrt(v0x + v0y)= sqrt(169 + 34.81)= 14.28 m/s

As for the launch angle is this calculation correct: α= (tan)-(v0y / v0x)= 24.49 ?

Please. let me be correct!
Correct.

AM
mmoadi
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#18
Oct15-09, 11:17 AM
P: 157
Thank you for your help, patience and time!


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