Projectile motion of a particle

In summary: 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.
  • #1
Nylex
552
2
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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  • #2
Looks good to me.
 
  • #3
Cheers Doc! :)
 
  • #4
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 :(.
 
  • #5
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 [itex]u cos\theta[/itex] (at the top of the motion)... then increases again back to the original value (when it's back to the starting height).
 
  • #6
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 [itex]u cos\theta[/itex] (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.
 

Related to Projectile motion of a particle

What is projectile motion?

Projectile motion is the motion of an object through the air or space under the influence of gravity. It follows a curved path known as a parabola.

What factors affect the projectile motion of a particle?

The factors that affect the projectile motion of a particle are the initial velocity, angle of projection, mass of the particle, and the force of gravity acting on the particle.

How is the trajectory of a projectile calculated?

The trajectory of a projectile can be calculated using the equations of motion, taking into account the initial velocity, angle of projection, and time elapsed. The trajectory will be a parabola.

What is the maximum height reached by a projectile?

The maximum height reached by a projectile is when the vertical component of its velocity becomes zero. This occurs at the highest point of the projectile's trajectory.

What is the range of a projectile?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the equation R = (v^2 * sin2θ)/g, where v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

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