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