calculate the angle of projection if its horizontal range is three times that of maximum height find the time of flight if its velocity of projection is 18 metre per second
Answers
The angle of projection = 53°
The time of flight of the projectile = 2.88 seconds
Given that the horizontal range of the projectile is equal to three the maximum height of the projectile.
Range = R = u² sin2θ /g
Height = h = u² sin²θ /2g
θ - the angle of projection
u - velocity of projection = 18 m/s
Given , R = 3h
=> u² sin2θ /g = 3 u² sin²θ /2g
=> sin2θ = 3 sin²θ/2
=> 2sinθ cosθ = 3sin²θ/2
=> 4/3 = tanθ
=> θ = 53°
The angle of projection = 53°
Time of flight of the projectile =
2u sinθ/g
= (2×18×(4/5))/10
= 2.88 seconds
Time of fight of projectile = 2.88 seconds
Answer:
The angle of projection = 53°
The time of flight of the projectile = 2.88 seconds
Given that the horizontal range of the projectile is equal to three the maximum height of the projectile.
Range = R = u² sin2θ /g
Height = h = u² sin²θ /2g
θ - the angle of projection
u - velocity of projection = 18 m/s
Given , R = 3h
=> u² sin2θ /g = 3 u² sin²θ /2g
=> sin2θ = 3 sin²θ/2
=> 2sinθ cosθ = 3sin²θ/2
=> 4/3 = tanθ
=> θ = 53°
The angle of projection = 53°
Time of flight of the projectile =
2u sinθ/g
= (2×18×(4/5))/10
= 2.88 seconds
Time of fight of projectile = 2.88 seconds