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# Mechanics 2 - a challenging question watch

1. I need help solving this Mechanics 2 question which is from the topic "Kinematics of a particle moving in a straight line or plane" and more specifically it has to do with Projectiles. Thanks in advance and here's the question:

A particle is projected from point O with speed u at an elevation α. If x and y are the horizontal and vertical displacements at time t, show that:
y = xtanα - gx^2(1 + (tanα)^2)/2u^2

At time t, the particle is at point P(x,y) and is moving upwards at an angle β. If OP makes an angle θ with the horizontal, show that:
2tanθ = tanα + tanβ. Hence deduce the direction of motion of a particle projected from O with speed 35 m/s at an elevation of 30° as it passes a point (15, 7.46).
2. (Original post by FMStudent)
I need help solving this Mechanics 2 question which is from the topic "Kinematics of a particle moving in a straight line or plane" and more specifically it has to do with Projectiles. Thanks in advance and here's the question:

A particle is projected from point O with speed u at an elevation α. If x and y are the horizontal and vertical displacements at time t, show that:
y = xtanα - gx^2(1 + (tanα)^2)/2u^2

At time t, the particle is at point P(x,y) and is moving upwards at an angle β. If OP makes an angle θ with the horizontal, show that:
2tanθ = tanα + tanβ. Hence deduce the direction of motion of a particle projected from O with speed 35 m/s at an elevation of 30° as it passes a point (15, 7.46).
So what have you done, and where are you stuck?
3. I tried using formulae
u - initial velocity
α - angle above horizontal
t - time

Using x = utcosα and y = utsinα - 0.5gt^2 I got y/x = tanα - gt^2/2utcosα => y = xtanα - xgt/ucosα => y = xtanα - gtx^2/ucosα.utcosα => y = xtanα - gx^2/u^2(cosα)^2
using that (secα)^2 = 1/(cosα)^2 and 1 + (tanα)^ = (secα)^2 I get:
y = xtanα - gx^2(1 + (tanα)^2)/u^2 and not y = xtanα - gx^2(1 + (tanα)^2)/2u^2 so I am missing the 2 on the bottom. Did I make a mistake somewhere.
I am a foreign student so I don't understand the second part at all so I would appreciate some help. Thanks
4. (Original post by FMStudent)
I tried using formulae
u - initial velocity
α - angle above horizontal
t - time

Using x = utcosα and y = utsinα - 0.5gt^2 I got y/x = tanα - gt^2/2utcosα =>
y = xtanα - xgt/**ucosα =>
y = xtanα - gtx^2/ucosα.utcosα =>
y = xtanα - gx^2/u^2(cosα)^2
using that (secα)^2 = 1/(cosα)^2 and 1 + (tanα)^ = (secα)^2 I get:
y = xtanα - gx^2(1 + (tanα)^2)/u^2 and not y = xtanα - gx^2(1 + (tanα)^2)/2u^2 so I am missing the 2 on the bottom. Did I make a mistake somewhere.
I am a foreign student so I don't understand the second part at all so I would appreciate some help. Thanks
I 've split it up as it's easier to read.

You dropped the 2 where I put the **. Other than that, it's fine.

For the second part.
"moving upwards at an angle β" means that the velocity makes an upwards angle of β with the horizontal.

"OP makes an angle θ with the horizontal". means that a line joining the point of projection to the point P where the particle is, makes an angle of θ with the horizontal (i.e. tan θ = y/x)

Does that clarify it?
5. ok, now I see where i made a mistake. Thanks. I understand the angles now but still have a problem with the last part - "Hence deduce the direction of motion of a particle projected from O with speed 35 m/s at an elevation of 30° as it passes a point (15, 7.46)." What exactly am I supposed to do here? sorry my english isn't very good.
6. (Original post by FMStudent)
ok, now I see where i made a mistake. Thanks. I understand the angles now but still have a problem with the last part - "Hence deduce the direction of motion of a particle projected from O with speed 35 m/s at an elevation of 30° as it passes a point (15, 7.46)." What exactly am I supposed to do here? sorry my english isn't very good.
When they say deduce, they mean find.

So you need to find the angle the direction of motion (the velocity, beta) makes at the point (15,7.46).

When they say "hence" is usually means this follows on from the last bit of work you did, so you are expected to used the formula "2tanθ = tanα + tanβ."

You are told what alpha is and can easily work out theta, and hence using that equation you can work out beta, the required angle.
7. Ok I think I got it now. Thanks a lot, you've been great help
8. You're welcome.
9. I think I ended up doing it wrongly as someone told me I didn't find what I should have and now I have no idea what I really have to do to solve the question. Can anyone please solve the last part of the question and show the working. Here's the question again:

A particle is projected from point O with speed u at an elevation α. If x and y are the horizontal and vertical displacements at time t, show that:
y = xtanα - gx^2(1 + (tanα)^2)/2u^2

At time t, the particle is at point P(x,y) and is moving upwards at an angle β. If OP makes an angle θ with the horizontal, show that:
2tanθ = tanα + tanβ. Hence deduce the direction of motion of a particle projected from O with speed 35 m/s at an elevation of 30° as it passes a point (15, 7.46).

I just need the part in bold. Thanks
10. (Original post by FMStudent)
At time t, the particle is at point P(x,y) and is moving upwards at an angle β. If OP makes an angle θ with the horizontal, show that:
2tanθ = tanα + tanβ. Hence deduce the direction of motion of a particle projected from O with speed 35 m/s at an elevation of 30° as it passes a point (15, 7.46).

I just need the part in bold. Thanks

Tan theta is just y/x, and you've just worked out a formula for y in terms of x.

So use that and multiply the whole lot by 2, to get 2 tan theta.

You also need to work out tan beta.

This is just the vertical velocity divided by the horizontal velocity. Eliminate "t" as you did for the first part of the question, and compare it with your equation for tan theta, and you will be able to substitute it in, to get the equation they are asking for.

If you're still stuck, post some working, and someone can check where you're going wrong.
11. now it's fine i did it

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