Projectile motion of a particle

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The discussion focuses on calculating the projectile motion of a particle launched with an initial velocity at an angle. The range, maximum altitude, and time to reach maximum altitude are derived using trigonometric components of the initial velocity. The range is expressed as X = (u^2.sin 2θ)/g, the maximum height as Y = (1/2g)(u sin θ)^2, and the time to reach maximum height as T = (u sin θ)/g. Concerns arise regarding the plot of velocity as a function of time, which appears as a straight line rather than the expected curve. Clarifications indicate that the velocity should start at its maximum, decrease to a minimum at the peak, and then return to the initial value.
Nylex
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Some particle is given an initial velocity, u at an angle θ to the horizontal. I'm asked to find (as a function of θ): 1. the range of the particle, X, 2. the maximum altitude reached, Y and the time taken to reach maximum altitude, T.

First I resolved u into components: uy = usin θ, ux = ucos θ.

For 1, I said x = ucos θ.t (since the horizontal motion is unaccelerated, no air resistance).

To work out t, I used s = ut + (1/2)at^2 for the vertical motion, setting s = 0. For t != 0, I got t = (2usin θ)/g

=> X = (ucos θ.2usin θ)/g = (2u^2.sin θcos θ)/g = (u^2.sin 2θ)/g

For 2, I used the fact that v = 0 when the particle reaches its maximum height and the equation v^2 = u^2 + 2as.

=> Y = (1/2g)(uy)^2 = (1/2g)(usin θ)^2

For 3, I again used v = 0 at maximum height, but used v = u + at

=> T = (usin θ)/g

Then, I'm asked to work out the projectile's velocity as a function of time.

v(t) = [(vx)^2 + (vy)^2]^1/2

vx = ucos θ
Using v = u + at, vy = usin θ - gt

v(t) = [(ucos θ)^2 + (usin θ - gt)^2]^1/2

v(t) = [(ucos θ)^2 + ((usin θ)^2 - 2gtusin θ + (gt)^2]^1/2

v(t) = [u^2.(cos^2 θ + sin^2 θ) - 2gtusin θ + (gt)^2]^1/2

v(t) = [u^2 - 2gtusin θ + (gt)^2]^1/2

Is this correct?
 
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Looks good to me.
 
Cheers Doc! :)
 
Hmm, I have to plot v(t) vs. t, for a given u and θ and all I get is a straight line. It doesn't seem right to me for some reason :(.
 
Nylex said:
Hmm, I have to plot v(t) vs. t, for a given u and θ and all I get is a straight line. It doesn't seem right to me for some reason :(.
Well, that can't be right. (I assume you are plotting the magnitude of the velocity.) Try it again! It starts out with its maximum value of u ... decreases to a minimum value of u cos\theta (at the top of the motion)... then increases again back to the original value (when it's back to the starting height).
 
Doc Al said:
Well, that can't be right. (I assume you are plotting the magnitude of the velocity.) Try it again! It starts out with its maximum value of u ... decreases to a minimum value of u cos\theta (at the top of the motion)... then increases again back to the original value (when it's back to the starting height).


Thanks again Doc. Yes, I'm plotting the magnitude of the velocity. My working for v is in my first post and I'm not sure what's wrong with it. Maybe I shouldn't have factorised u^2.cos^2 θ + u^2.sin^2 θ by u^2, but it shouldn't make a difference.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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