Two particles are simultaneously thrown in horizontal direction from two points on a riverbank, which are at certain height above the water surface. The initial velocities of the particles are `v_1= 5m//s and v_2 = 7.5 m//s` respectively. Both particles fall into the water at the same time. First particles enters the water at a point s = 10 m from the bank. Determine
(a) the time of flight of the two particles,
(b) the height from which they are thrown,
(c) the point where the second particle falls in water.

Options

A. `v^(2)/(2g)`
B. `v^(2)/g`
C. `(4v^(2))/g`
D. `(2v^(2))/g`

Answer

C

Explanation

Here in this case, for particle projected along +ve x-axis, position vector at time t.
`vecr_(1) = vt hati - 1/2gt^(2) hatj`
Again, for other particle which is projected along -ve x-axis, position vector at time t.

`vecr_(2) = vt(-hati) -1/2 gt^(2) hatj`
Now, as per question,
`(vecr_(1) bot vecr_(2)), therefore vecr_(1).vecr_(2) =0`
`rArr -v^(2)t^(2) + 1/4 g^(2)t^(4)=0 rArr v^(2)t^(2) = 1/4 g^(2) t^(4)`
`t= (2v)/g`
Therefore, separation between the particles at time .t. is
`Deltax = vt -(-vt) = 2vt = 2 xx v xx (2v)/g = (4v^(2))/g`