The Maths and Science Thread - Collection of Problems and Facts

[Varun]

Suppose sinA = sinB = 1, then cos A = cos B = 0 (A = B = 90degrees). But then you could have sinC = -1, and cos C = 0 (C = 270degrees)

Then sinA + sinB + sinC = 1 and cosA + cosB + cosC = 0

Are there more possibilities?

 

[Dr. Pepper]

Okay I'll go through these questions and get to the last one which I'm stuck on:

An empty car rolls down a slope and off a cliff edge which is 20m vertically high. It leaves the cliff top travelling 25m/s horizontally.

1a) Calculate the time it would take for the car to reach the beach at the bottom of the cliff.

Here's my calculations:

s=ut + 1/2 at^2
u is 0 so it goes away
2s=at^2
t^2=2s/a
t= the square root of 2s/a
which is = the square root of (2x20)/9.81
= 2.02 seconds

1b) Calculate how far from the bottom of the cliff it would land in the sand

s/t= u+v/2
s=u+v/2 xt
=25+0/2 x2.02
= 25.25m

1c) The vertical speed it would have just as it hits the sand

v= u - at
v= 0 - 9.81 x 2.02
= 19.8m/s

1d) work out the diagonal velocity it'd have when it lands (with the angle)

I'm not too sure about 1c) and I have no idea what to do for 1d). Can someone please help. Also if you think I did anything else wrong please say.

 

[Varun]

Wait, isn't this projectile motion? We cannot simply use the equations straightaway, we should apply trigo in that, right?

 

[Dr. Pepper]

Yeh it is projectile motion.

We went over it in class and basically I just tried to over-complicate things when it was actually really basic.

1c) was just s=v x t which was equal to 25x2.02= 50.5

And d) you just needed to use pythag to work out the velocity. Diagonal velocity = the square root of 19.81^2 + 25^2. Then for the angle use trig from that.

 

[Kshitiz_Indian]

Okay I might not be getting you correctly, but even in Projectile motion, you can apply the normal equations in ONE axis. So you can apply the normal equation in relation to the vertical velocity.

In this case, if I'm right, the vertical velocity will initially be 0. So you just use that equation with initial velocity 0. That's right.

The horizontal velocity will NOT change during this whole period, if we dis regard air friction / viscosity.

The diagnol velocity at the bottom will be the vector sum of the horizontal and the vertical velocity at that point. You know Vectors?

 

[Dr. Pepper]

Yeh so if it's a constant horizontal velocity then I can use s=v x t, right?

Not sure what you mean by vectors though, but I just used pythag because I have two of the sides and I just need the hypotenuse.

 

[Kshitiz_Indian]

Yeh so if it's a constant horizontal velocity then I can use s=v x t, right?

Right. Use the t that you evaluated from the vertical equation and you'll get the distance it travels horizontally.

Not sure what you mean by vectors though, but I just used pythag because I have two of the sides and I just need the hypotenuse.

The Pythagoras thing comes from a concept known as Vectors, which I guess you'll study later if you opt for higher studies in Physics. Forget it now anyways.

Pythagoras theorem will do the trick, yeah!

 

[Varun]

Lol the only thing I know about a vector quantity is that it has magnitude as well as direction. Some of their rules are similar to the coordinate geometry ones. Nothing else.

 

[AkshayC]

If i dig a hole through the Earth (from one end to another), then if i jump through it, will i fall out the other side or what? Because when i jump in it i will be falling DOWN but how could i come out the other side falling DOWN. Thats impossible. :help:help:help:help:help

 

[P Squared]

Digging a hole from end to another is impossible. This situation will never arise.

 

[AkshayC]

Digging a hole from end to another is impossible. This situation will never arise.

but how ??

 

[Kshitiz_Indian]

I think if you jump in that you'll start oscillating between the two poles of the hole. Not so sure, will have to check on that, but my memory tells me about the oscillation theory.

 

[Varun]

Actually I kinda agree with Kshitiz,

We know that the earth's gravitational pull decreases as we reach the core.
So when we'll move inside the earth, suppose we first reach to some part of the core (which we certainly will), and then we keep falling downwards (going ahead of the core), indeed the earth's gravitational pull will increase and it will pull us back to the core. Again after passing the core, the gravity will increase, then we'll be pulled back towards the core.

This way we'll keep oscillating between the holes.
I hope you get it.

 

[AkshayC]

Is it correct that rather than coming out from other side will more or less stay in middle because I will be pulled by gravity equaly in all directions by a force of 1G.
effectively I have as much Earth mass pulling you down as trying to pull you

 

[Varun]

(See the image from bottom to top.)

What my observations say is that we'll keep oscillating between the two ends of hole passing through the core.