#1
Oct105, 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 . 



#2
Oct105, 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.



#3
Oct205, 09:55 AM

P: n/a

Which equation for motion?
I still do not understand. 



#4
Oct205, 10:36 AM

P: 226

Projectile Motion[tex] v_{y1} = v_ {y0} + g't \Longrightarrow g' = \frac{v_{y1}v_{y0}}{t} [/tex] You could then put v_{y2} and v_{y3} in terms of v_{y0}, 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. 


#5
Oct205, 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? 



#6
Oct205, 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?



#7
Oct205, 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. 



#8
Oct205, 11:24 AM

HW Helper
P: 482

[tex]v_y = v_{y0}  gt[/tex]
What do you know at t = 1s and t = 2s? 


#9
Oct205, 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 



#10
Oct205, 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. 



#11
Oct205, 11:33 AM

P: 226

This answer is correct; however, by looking at your graph I can tell you the *exact* numerical value of v_{2y}. Why is that? 


#12
Oct205, 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. 


#13
Oct205, 12:07 PM

P: n/a

Any responses?




#14
Oct205, 12:18 PM

HW Helper
P: 482

You have radians on.
Looks good. 


#15
Oct205, 10:15 PM

P: n/a

Thank you everyone.



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