Question

A particle is dropped from a height h. Another

particle which is initially at a horizontal distance

‘d’ from the first is simultaneously projected with

a horizontal velocity u and the two particles just

collide on the ground. Then :

(a)

2

2 u h d

2g

(b)

2

2 2u h d

g

(c) d = h (d)

2 2 gd u h​

Answer 1

Answer:

HOPE IT HELPS U BUDDY.....

Answer 2

Answer:

$T = \sqrt{\frac{2h}{g} }$

therefore:-

$D = ut$

$d = u\sqrt{\frac{2h}{g} }$

$d^{2} = 2u^{2} h/g$

Thank you....