Projectile Motion Question - Not enough info?

[Sean1218]

Homework Statement

http://oi51.tinypic.com/9jekhu.jpg

The ball is at the bottom of the ramp, and you push it (one push to launch it) up the ramp, through the photogate, through the air to hit the last platform.

\Deltad_x = 1.00 m
\Deltad_y = 0 m (because the two platforms are level
Diameter of the ball = 0.045 m
Time it takes the ball to go through the photogate = 0.086 s
g = -9.8 m/s²

1) Determine the initial velocity of the ball.
2) Determine the length of time that the ball was in the air.

Homework Equations

Uniform acceleration equations
V = \Deltad / \Deltat
V_1x = V_2x
V_1x = V₁cosθ
V_1y = V₁sinθ
V₁ = sqrt(V_1x² + V_1y²)

The Attempt at a Solution

Ok, the first thing I thought to do was divide the diameter of the ball by the time it takes the ball to go through the photogate.

0.045 m / 0.086 s = 0.52 m/s

However, I can't figure out whether 0.52 m/s is the value for V_1x or if it's the value for V₁ (in which case I'm not given the angle).

There's nothing I can do with it if it's equal to V₁, so I'm assuming V_1x = 0.52 m/s

I tried everything I could think of in both the x-direction (nothing you can do here) and y-direction (I can't find the right equation, it feels like I don't have enough info to do anything with it). V_1x = V_2x, but I can't find anyway to make use of the V_2x, because I need it in terms of y to do anything with it.

I've been trying to do this problem about an hour and a half, but it seems to be outside of my abilities.

Any guidance?

 

[ehild]

Make a drawing of the ball entering and leaving the photogate, and see the components of the displacement of the CM to make sure what you get when dividing the diameter of the ball with the time.

If you are sure which component of the velocity you got then use the relation between the x and y coordinates of the projectile.

ehild

 

[Delphi51]

I think your .52 m/s is the initial velocity of the ball, the answer to part 1. Try doing the projectile motion, horizontal and vertical parts. I think you will find you have only 2 unknowns θ and t so you can solve for them using the system of 2 equations (one horizontal and one vertical).

 

[OOOl.lo]

Mr Le's Class?

 

[ehild]

I think your .52 m/s is the initial velocity of the ball, the answer to part 1.

Think it over...

ehild

 

[Delphi51]

Okay, the vertical line for the photogate does suggest that it measures the horizontal time so .52 could be the Vx. I don't see what the center of mass position has to do with it . . .
The projectile motion part can be solved with either take on the .52.
Am I missing something obvious again?

 

[ehild]

Okay, the vertical line for the photogate does suggest that it measures the horizontal time so .52 could be the Vx. I don't see what the center of mass position has to do with it . . .
The projectile motion part can be solved with either take on the .52.
Am I missing something obvious again?

The motion of the projectile is considered as if the whole mass was concentrated in the CM. So you need the velocity of the CM.

The diameter over the time in the photogate is the horizontal component of the initial velocity. From here, you can solve the problem using the equations for the projectile. As the change of x and y are known, a straightforward way is to use the equation Δy with respect to Δx to get the y component of the initial velocity.
Δy=(v_1y/v_1x)Δx - g/2 (Δx/v_1x )².

ehild

 

[Sean1218]

So, we've established that V_1x = 0.52 m/s, but how would I go about answering the questions using the equations I listed?

I'm not exactly sure what equation you used (I don't think we've learned that yet), and I'm not sure if you mean change in V or change in d by Δy and Δx.

edit: resembles d = (v2)(t) - 1/2(g)(t)^2, but I'm still not quite sure what you did

 

[ehild]

You certainly learned that the motion of the projectile consist of a horizontal motion with constant velocity v_1x, and a vertical motion with an initial upward velocity v_1y. Write the equation for the horizontal and vertical displacements in terms of time, solve for the time till the ball reaches to the other side, and use this time to get v_1y. If you know both components of the initial velocity, you can find its magnitude.

ehild

 

[Sean1218]

I actually don't know how I didn't see this before. Read your last response and it just clicked, thanks!

 

[Sean1218]

After thinking it over and talking to my teacher, apparently the diameter of the ball over the time it takes it to go through the photogate = V₁, not V_1x.

Also, apparently we have to use some trig identities to solve it.

Comments?

 

[ehild]

Why should it be v1?
The photogate is vertical. The ball enters it when the centre of the ball is at distance R from the light ray of the LED and leaves it when the centre is at the other side at distance R. See the picture in my former post. So the horizontal component of displacement is 2R during the time the ball is in the gate.

ehild