A particle is projected at an angle 30° to the horizontal and 2 seconds later is moving in the direction arctan(1/4) to the horizontal. Find its initial speed

Question

Mechanics.

Answer 1

A particle is projected at an angle 30° to the horizontal and 2 seconds later is moving in the direction arctan(1/4) to the horizontal. Find its initial speed

Let V be the initial speed at 30° projection to horizontal.

Horizontal velocity= Vx = V.cos 30  (this velocity is constant)

Vertical velocity = Vy = V.sin 30       (this velocity is affected by gravity)

Using v = u + at,

After 2 secs, Vy2 = Vy – 9.81*2

Vy2 = V.sin 30 – 19.62

Vy2 = V/2 – 19.62

Vx2 = the same (constant) velocity the particle started with.

Vx2 = V.cos 30

Vx2 = sqrt(3)V/2

Angle of flight is given by tan α = Vertical velocity / horizontal velocity

tan α = Vy2 / Vx2

tan α = (V/2 – 19.62) / sqrt(3)V/2  

tan α = 1/sqrt(3) – 39.24/sqrt(3)V

¼  = 1/sqrt(3) – 39.24/sqrt(3)V

39.24/sqrt(3)V = 1/sqrt(3) – ¼ = 0.3274

sqrt(3)V = 39.24 / 0.3274 = 119.87

V = 119.87 / sqrt(3) = 69.2

V = 69 m/s