Question

A particle is projected with velocity $2\sqrt{gh},$ so that it just clears two walls of equal height h which are at a distance of 2h from each other. Find the time of passing between the walls.

• A. $\sqrt{\frac{h}{g}}$
  
• B. $2\sqrt{\frac{h}{g}}$
• C. $2\sqrt{\frac{g}{h}}$
• D. $3\sqrt{\frac{g}{h}}$

Answer 1

Answer: (B)

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

u = 2[√(gh)]
horizontal speed = ucosθ = 2cosθ√(gh)
therefore time to cover the interwall distance = t = 2h/[2cosθ√(gh)] = √(h/g) / cosθ
vertical velocity² at the top of ist wall = v² = 4ghsin²θ - 2gh = 2gh[2sin²θ - 1]
v = √{2gh[2sin²θ - 1]}
t = 2v/g = 2√{2gh[2sin²θ - 1]} / g = √(h/g) / cosθ -------------------------> (i)
squaring => 4{2gh[2sin²θ - 1]}cos²θ = hg
=> 8[1 - 2cos²θ]cos²θ = 1
or -2cos⁴θ + cos²θ - (1/8) = 0
=> solution is cosθ = 0.5
or θ = 60°
=> from equation (i) => t = 2√(h/g)