Projectile Motion (1 Viewer)

[Aerath]

Hey guys, just wondering if some people could help me out with a couple of questions:

1. A ball is dropped from a lookout 320m high. At the same time, a stone is fired vertically upwards from the valley floor with a speed of v m/s (g = 10m/s^2).
a) Show that if the ball and stone collide in the air, they do so when V is greater or equal to 40m/s.
b) [I got b]

2. Two particles are projected simultaneously from two points A and B on level ground and a distance of 150m apart. The first particle is projected vertically upwards from A with an initial speed of u m/s and the second particle is projected from B towards A with an angle of projection alpha. If the particles collide when they are both at their greatest height about the level of AB, prove that:
tan alpha = (u^2) / (150g)

3. The nozzle of a water hose is at point O on the horizontal ground. The water comes out of the nozzle with a speed U m/s. Neglecting the air resistance, prove that the water can reach the wall at a distance "d" from O, if U^2 > gd, when g is acceleration due to gravity. If U^2 = 4gd, also prove that the maximum height that can be reached by the jet on this wall is given by 15d/8.


Any help on these questions would be greatly appreciated.

 

[ronnknee]

 

[lyounamu]

Damn, didn't even finish the question. And this guy took the only question I probably knew.

But that's 100 times better than I was about to change to.

Can I quickly ask you a question? I thought if the one had the positive acceration, the other should have negative acceleration. That's the only flaw I made. Can you tell me why they both have same sign for the acceleration?

 

[Aerath]

Because acceleration is always negative, as gravity always goes towards earth.

 

[lyounamu]

Because acceleration is always negative, as gravity always goes towards earth.

I knew that but doesn't it differ when it comes to the different direction?

If you treat the upwards being positive, the downwards must be negative and vice versa. That's how I got taught by my teacher.

 

[Aerath]

Yeah g = 10m/s^2, but y'' = -g, therefore acceleration of y is -10m/s^2.
[At least I think that's what you meant.]

 

[lyounamu]

Yeah g = 10m/s^2, but y'' = -g, therefore acceleration of y is -10m/s^2.
[At least I think that's what you meant.]

Yep. I am starting Projectile Motion now. There are tonnes of formulas to derive and memorise. Not my type of topic. :mad1:

 

[vds700]

Yep. I am starting Projectile Motion now. There are tonnes of formulas to derive and memorise. Not my type of topic. :mad1:

Namu u need to derive them in exams, not memorize them and just state the formulas. Also, many questions have special conditions like a stone thrown off a cliff which the standard projectile motion equations do not apply.

 

[u-borat]

I knew that but doesn't it differ when it comes to the different direction?

If you treat the upwards being positive, the downwards must be negative and vice versa. That's how I got taught by my teacher.

You only do that in 4 unit where you separate it into 2 components.
I'm pretty sure you can always take acceleration to be negative 9.8 in 3 unit projectiles...

 

[Yip]

2.
Equation of motion for A: dy/dt=-gt+u (where y represents horizontal displacement)
A reaches its greatest height when dy/dt=0
Denoting time of collision by T, T=u/g

Equation of motion for B: dy/dt=-gt+vsin(a) (a is the angle of projection alpha, v is
the velocity of projection from B)
B reaches its greatest height when dy/dt=0
Since A and B reach their respective greatest heights at collision,
u/g=vsin(a)/g, i.e. v=u/sin(a)

x=vTcos(a)=150 (since A and B collide 150m away from the projection point of B)
vTcos(a)=[u/sin(a)][u/g]cos(a)=u^2/gtan(a)=150
tan(a)=u^2/150g

3. Equation of motion for water:
x=utcos(a)
y=-0.5gt^2+utsin(a)
First step is to derive an expression for the horizontal range of the water,
which can be done by setting y=0
usin(a)=0.5gt
t=2usin(a)/g
Substituting into x to find the horizontal range H,
H=u^2[2sin(a)cos(a)]/g=u^2sin(2a)/g
Since the water has to touch the wall, H>d
i.e. u^2sin(2a)/g>d
Since sin(2a)<=1 for all a,
u^2/g>=u^2sin(2a)/g>d
i.e. u^2>gd

If u^2=4gd, we need to show that the highest point the water can touch the wall is 15d/8

x=utcos(a)=d
Set u=2(gd)^0.5,
t=[(d/g)^0.5]/[2cos(a)]
Setting thie value of t into y to get the height,
Height=-0.5g(d/4g[[cos(a)]^2])+[2(gd)^0.5][sin(a)][(d/g)^0.5]/[2cos(a)]
=(-d/8)[sec(a)]^2+dtan(a)=(-d/8)[[tan(a)]^2+1]+dtan(a)
=(-d/8)[[tan(a)]^2-8tan(a)+1]
We want to find the value of tan(a) such that the height is maximized.
The expression in the brackets is a quadratic in tan(a), so the extreme value occurs at tan(a)=4.
Setting tan(a)=4,
Height=(-d/8)(-15)=15d/8

 

[lyounamu]

Namu u need to derive them in exams, not memorize them and just state the formulas. Also, many questions have special conditions like a stone thrown off a cliff which the standard projectile motion equations do not apply.

Thanks for the tips. In my class, we haven't even started SHM and Projectile Motion. Now, I am forced to study this myself. (and no one in my class probably does this)

It's so stupid that we are learning this just before the Trials. Trust me, everyone except few in my class will fail the Trials due to this.

Good luck with your studying by the way.

 

[lyounamu]

You only do that in 4 unit where you separate it into 2 components.
I'm pretty sure you can always take acceleration to be negative 9.8 in 3 unit projectiles...

Oh. Thanks for clarifying that up.

 

[vds700]

Thanks for the tips. In my class, we haven't even started SHM and Projectile Motion. Now, I am forced to study this myself. (and no one in my class probably does this)

It's so stupid that we are learning this just before the Trials. Trust me, everyone except few in my class will fail the Trials due to this.

Good luck with your studying by the way.

ah you'll be fine. We're screwed in 4 unit, we haven't even started volumes and harder 3 unit (only done circle geometry), its gonna be such a cram before the trials

 

[ronnknee]

Can I quickly ask you a question? I thought if the one had the positive acceration, the other should have negative acceleration. That's the only flaw I made. Can you tell me why they both have same sign for the acceleration?

Acceleration is ALWAYS gravity acting towards the ground, when near the surface. Although it varies as altitude varies in real life, we treat the acceleration of gravity as constant

What you are thinking of is the velocity. When the arrow is launched from the ground, it is still affected by gravity ie. being pushed down, while it tries to go up. When the initial velocity is less than gt, the final velocity becomes negative and the arrow falls back down

 

[lyounamu]

ah you'll be fine. We're screwed in 4 unit, we haven't even started volumes and harder 3 unit (only done circle geometry), its gonna be such a cram before the trials

LOL. But you guys are at least more academically adept than we are.

 

[lyounamu]

Acceleration is ALWAYS gravity acting towards the ground, when near the surface. Although it varies as altitude varies in real life, we treat the acceleration of gravity as constant

What you are thinking of is the velocity. When the arrow is launched from the ground, it is still affected by gravity ie. being pushed down, while it tries to go up. When the initial velocity is less than gt, the final velocity becomes negative and the arrow falls back down

Ok. My brain got screwed due to excessive cramming.

 

[ronnknee]

 

[Yip]

Ronnknee, your working is incorrect. sin(a) cannot =1, as that implies a=pi/2, which implies a vertical projection, which would result in the 2 particles never colliding. If a were to be pi/2, tan(a) would then be undefined anyway. Also u doesnt necessarily have to equal v. Refer to the solution i posted on the previous page for a more mathematically correct solution.

 

[ronnknee]

Now that you pointed it out, yea I can see what's incorrect

I was thinking, at a certain time their heights and vertical velocities are equivalant. Wouldn't that be true?

 

[Yip]

The crux of this question is to realize that at collision, the 2 particles will have the same height. Then u want to find a relation between the projection speed of B, which is v, and the projection speed of A (btw thats another flaw of ur previous attempt, u is not the projection speed of B). This can be done by first deriving the time at which particle A and particle B reach their respective maximum heights. After finding these times for A and B, u equate them. The rest should follow pretty easily.