Question

velocity of a projectile is v at half of its maximum height and it is root 2 upon 5 v at maximum height find the angle formed by the projectile​

Answer 1

Answer:

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