Total speed of a projectile at its greatest height is root 6 root 7 of its speed when it is at half of its greatest height
Answers
Answer:
Angle of Projection = 30°
Step-by-step explanation:
Total speed of a projectile at its greatest height is root 6 root 7 of its speed when it is at half of its greatest height
let say initial Speed of Projectile = V
Horizontal speed = Vcosα
Vertical Speed = Vsinα
Speed at greatest height = Vcosα ( as Vertical speed = 0)
Using V²-U² = 2aS
Height H = V²sin²α/2g
half of its greatest height = V²sin²α/4g
Vertical velocity at half of its greatest height = Vₓ
V²sin²α/4g = (V²sin²α -Vₓ²)/2g
=> V²sin²α = 2V²sin²α -2Vₓ²
=> Vₓ² = V²sin²α/2
=> Vₓ = Vsinα/√2
Horizontal Velocity at half of its greatest height = Vcosα
Total Velocity = √(Vsinα/√2)² + (Vcosα)²
= √(V²sin²α/2 + V²cos²α)
Vcosα/√(V²sin²α/2 + V²cos²α) = √6/√7
Squaring both sides
V²cos²α/(V²sin²α/2 + V²cos²α) = 6/7
=> 7V²cos²α = 6V²sin²α/2 + V²cos²α
=> V²cos²α = 3V²sin²α
=> cos²α = 3sin²α
=> 1/3 = tan²α
=> tanα = 1/√3
=> α = 30°
Angle of Projection = 30°