Projectile Motion


by dekoi
Tags: motion, projectile
dekoi
#1
Oct1-05, 12:49 PM
P: n/a
If the velocity of a projectile is known at t=1second and the gravitational force is unknown, how does one determine the velocity at t=0, 2, and 3 seconds.

The projectile does not start from rest.


If i use the formula: vfy = viy - gt, i can't solve for v, since the value for g is unknown.

Thank you .
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hotvette
hotvette is offline
#2
Oct1-05, 08:52 PM
HW Helper
P: 930
Use the information you have. How can knowing the velocity at t=1 sec help? Hint: use your equation of motion.
dekoi
#3
Oct2-05, 09:55 AM
P: n/a
Which equation for motion?

I still do not understand.

Grogs
Grogs is offline
#4
Oct2-05, 10:36 AM
P: 226

Projectile Motion


Quote Quote by dekoi
The projectile does not start from rest.
You're sure this is correct and you've listed everything that's been given to you (vy1)? If that's the case, you can't solve for vy0, vy2, etc. numerically. The best you could do is solve for them in terms of vy0 . For example:

[tex]
v_{y1} = v_ {y0} + g't \Longrightarrow g' = \frac{v_{y1}-v_{y0}}{t}
[/tex]

You could then put vy2 and vy3 in terms of vy0, but that's as far as you could go. If you had one other piece of information you could solve for g' numerically and use it to solve for the other velocities as well.
dekoi
#5
Oct2-05, 10:57 AM
P: n/a
The only problem with that is that in the second part of the question, I have to find the value of "g", so there must be a way to do it. Here is the exact question:

A physics student on Planet Exidor throws a ball, and it follows a parabolic trajectory. The ball's position is shown at 1 s intervals until t = 3 s. At t = 1 s, the ball's velocity is v = (2.0i + 2.0j) m/s. (i and j being unit vectors for x and y)
a) Determine the ball's velocity at t = 0s, 2s, and 3s.
b) What is the value of g on Planet Exidor?
c) What was the ball's launch angle?
Päällikkö
Päällikkö is offline
#6
Oct2-05, 11:12 AM
HW Helper
P: 482
Does the problem have a graph/figure ("The ball's position is shown" implies it does)? Can you see the height from that?
dekoi
#7
Oct2-05, 11:20 AM
P: n/a
It does not land at 3 seconds; the graph only shows up until 3 seconds.

The graph is attached.
Attached Thumbnails
graph.jpg  
Päällikkö
Päällikkö is offline
#8
Oct2-05, 11:24 AM
HW Helper
P: 482
[tex]v_y = v_{y0} - gt[/tex]
What do you know at t = 1s and t = 2s?
dekoi
#9
Oct2-05, 11:27 AM
P: n/a
At t = 1s, vx = 2.0 m/s and vy = 2.0 m/s.
At t = 2s, vx = 2.0 m/s and vy = (2.0 - g) m/s
Fermat
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#10
Oct2-05, 11:30 AM
HW Helper
P: 879
From your graph, you can see that the total time of flight is 4s.
So the time to reach max height = 2s
You also know the vertical velocity at t = 1.
By symmetry the velocity at t = 3 will be the same as at t= 1, but in the opposite direction.
Now you can use your eqns of motion to solve for g, etc.
Grogs
Grogs is offline
#11
Oct2-05, 11:33 AM
P: 226
Quote Quote by dekoi
At t = 2s, vy = (2.0 - g) m/s
{quote snipped for clarity}

This answer is correct; however, by looking at your graph I can tell you the *exact* numerical value of v2y. Why is that?
dekoi
#12
Oct2-05, 11:46 AM
P: n/a
Is this correct?

v2y= 0 m/s
therefore
0 m/s = 2.0 - g
g = 2.0 m/s2

v0 = (2.0i + 4.0j) m/s
v1 = (2.0i + 2.0j) m/s
v2 = (2.0i) m/s
v3 = (2.0i - 2.0j) m/s

v0 = sqrt(2.0^2 + 4.0^2) = sqrt(4 + 16) = sqrt20 m/s

angle = arccos(v0x / v0) = arccos(2 / sqrt20) = 1.1 degrees

I am not sure about the angle.
dekoi
#13
Oct2-05, 12:07 PM
P: n/a
Any responses?
Päällikkö
Päällikkö is offline
#14
Oct2-05, 12:18 PM
HW Helper
P: 482
You have radians on.

Looks good.
dekoi
#15
Oct2-05, 10:15 PM
P: n/a
Thank you everyone.


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