• A particle at a height h from the ground is projected with an angle of 30deg from the horizontal , it strikes the ground making an angle of 45deg with horizontal. It is again projected from the same point at height h with the same speed but with an angle of 60deg with the horizontal find  the angle it makes with the horizontal when it strikes the ground .

@vijay.

that is what i was getting too.

but i doubt it.

Let the body be projected from a height h from a point O above the ground. Its initial velocity is u then, and its angle of projection is 30 degrees. When it crosses the same height after the projection, its vertical velocity is –usin30 and the horizontal velocity ucos30 (which remains constant throughout the motion).

From here the body moves downwards with an initial vertical velocity of –usin30 and a constant horizontal velocity of ucos30. When it strikes the ground let the velocity be v.

And it makes an angle 45 degrees with the horizontal.

So, ucos30 = vcos45 …….(1)

Or, v = u*sqrt[3/2].

Also if we analyze the vertical motion using the third equation we get,

(v^≥2)/2 = (u^2)/4 – 2g(-h)

Or, 3(u^2)/4 = (u^2)/4 + 2gh.   Using equation (1)

Or, u  = sqrt[4gh]. ….(2)

Now we consider the second case similarly,

Here we let the final velocity when the body strikes the ground to be V.

So, if x is the angle made with the horizontal when it strikes the ground,

Vcosx = ucos60

Or, cosx = (u/2V) ……(3)

Now we consider the vertical motion of the body,

(Vsinx)^2 = 3(u^2)/4 – 2g(-h) ….(4)

Or, V^2 – (Vcosx)^2 = 3(4gh)/4 + 2gh  using equation (2).

 Or, V^2 – (u/2)^2 = 5gh

Or, V^2 = 6gh.

So, cosx = ½*sqrt[4gh]/sqrt[6gh]

Or, cosx = sqrt[1/6]

i don't know if i am correct or not....

let u be the velocity of the projection of the body,

then at the point of projection the total mechanical energy of the particle is 1/2 mu^2 + mgh

let v be the velocity at lowest point then 1/2 mv^2 = 1/2 mu^2 + mgh

v^2 = u^2 +2gh

also we know that horizontal velocity remains constant , so ucos30 = vcos45,

sqrt ( u^2 +2gh) / sqrt 2 = u*sqrt(3) /2

3u^2 / 2 = u^2 +2gh ,

u =2 * sqrt(gh)

let $ be the angle it makes with horizontal

again the total velocity remains the same , and the horizontal velocity remains the same

u cos60 = v cos $

u / sqrt ( u^2 + 2gh )  * 1/2 = cos $

2*sqrt(gh) / sqrt ( 6gh)  * 1/2 = cos $

cos $ = 1/ sqrt (6) ,

so initially i got wrong answer but now this one's correct :)