Question

A particle is projected with velocity 2√(gh) ,so that it just clears two wall  of equal height h which are at a distance of 2h from each other.show that the time of passing between the walls is ​

Answer 1

Explanation:

Correct option is

h g B 2₁

horizontal speed = ucos0 = 2cos0√(gh)

therefore time to cover the interwall

u = 2[√(gh)]

distance = t = 2h/[2cos0√(gh)] = √(h/g) / cose vertical velocity² at the top of ist wall = v² = 4ghsin²0 - 2gh = 2gh[2sin²0 - 1]

v = √{2gh[2sin²0 - 1]}

t = 2v/g = 2√{2gh[2sin²0 - 1]} / g = √(h/g) / (i)

cose.

squaring => 4{2gh[2sin²0 - 1]}cos²0 = hg

=> 8 [1-2cos²0]cos²0 = 1

or - 2cos^4 theta + cos^2 theta - (1/8) = 0

=> solution is cos theta = 0.5

or 0 = 60°

=> from equation (i) => t = 2√(h/g)