Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

thank you so much bob

anyhow, manually.

Did you appear in KVPY @agnishom.

**niharika_kumar**- Replies: 38

Find the value of

thank you so much bob.

bob bundy wrote:

Now for the horizontal motion. At impact with the block, the velocity is still V and the block is moving at -V/4. We aren't told the mass of the block so no momentum equation is possible. But we can use the elasticity equation. We'll have to assume e=1 again as we aren't told anything else. So

You'll need to be careful with negatives here, but you should be able to get the velocity of the ball after the impact (assume block is still -v/4).

Bob

I am not sure if I could understand the way of using that equation properly.

Will you pls show how you placed all the terms in that.

**niharika_kumar**- Replies: 5

The block A begin to move with a velocity v/4.At the same time a ball is thrown from the wall of height h with a velocity v which traces the path as shown before coming back to the original position.Find x.

Welcome to the forum!

here is one more:-

Show that all the chords of the curve

which subtend a right angle at the origin pass through a fixed point.Also find that point.my thanks to him too.

Even I was unable to do that and hence my 2nd equation was not in latex form.

thanks bob.

I didn't consider fact that y-mx is the factor of both the equations, but yes you used it, so

if we have that (y-mx) is the factor of both the given equations,

why don't we put y=mx in both of them, then eliminate the term x^2 and solve by cross multiplication.

that gives m^2 =(hA-Ha)/(bH-hB)={(aB-Ab)/2(bH-Bh)}^2

That has given a clear solution without m,n and p and the steps are less complicated.

Here is another question:-

Obtain the condition that one of the straight lines given by the equation

may coincide with one of those given by the equation

a'x^2+2h'xy+b'y^2=0

thank you bob.

Welcome to the forum!

h is the height to which the ball is raised up from its hanging position.

**niharika_kumar**- Replies: 4

In the given figure the ball of mass m is held above as shown.It is then allowed to swing freely.Find the maximum height attained by the block.

Yes you are.

Your explainations have always proved to be very beneficial.

hmm, so it was simple.

thanks again bob.

It's over now.

and I couldn't make this one:-

Q.A ball rolls off the top of a stairway with a horizontal velocity u.If the steps are h meter high and b meter wide, show that the ball will just hit the edge of nth step if

It is really very helpful(because my brain is not working right now and I have my Physics exam tomorrow).

**niharika_kumar**- Replies: 14

A particle is projected over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base.If

and be the base angle and the angle of projection , prove that .Welcome to the forum!

thanks bob.

no I couldn't solve it and so put it up here.

are you sure x=a and not -a.

I am getting x=-a.