Question

If the horizontal velocity of a projectile is root(2/5) times the velocity at half the maximum
height, the angle of projection is

Answer 1

horizontal velocity = ucosø
velocity at maximum height is zero
but there will be some velocity as it says half the maximum height
h = u²sin²ø/2g. then
h' = u²sin²ø/4g

now ,
v'²=u²- 2gh'
keeping u = vsinø
on solving this equation u will get velocity at half the maximum height I.e v'=usinø/√2

ucosø=√2/√5(u usinø/√2)
cosø=1/√5sinø
tanø=√5
ø=tan^-1{√5}

Answer 2

let ∅ is angle of projection , and u is initial velocity .
according to question ,
velocity of projectile in horizontal =√(2/5) velocity of projectile at half of maximum height .

we know,
velocity of projectile in horizontal =ucos∅ ------------(1)

we also know,
Hmax = u²sin²∅/2g

velocity at Hmax/2 .

use kinematics equation for Y-axis
Vy² =Uy² + 2ays
Uy = usin∅
sy =Hmax/2 = u²sin²∅/4g
a = -g
Vy² =( usin∅)² -2g{u²sin²∅/4g}

=u²sin²∅ -u²sin²∅/2

=u²sin²∅/2g

Vy = usin∅/√2g

now velocity in x -axis at Hmax/2
Vx = Ux +ax t
Vx =ucos∅
ax = 0
so, Vx =ucos∅

hence, velocity in vector form
V =ucos∅ i + usin∅/√2 j

|V| =u√(cos²∅ +sin²∅/2) ---------(1)



now,
ucos∅ =√(2/5)u√(2cos²∅ +sin²∅)/√2

cos²∅ = 1/5(2cos²∅ +sin²∅)

5cos²∅ = 2cos²∅ +sin²∅

3cos²∅ = sin²∅

tan²∅ = 3

tan∅= ±√3

so, ∅ = π/3 or , 2π/3