The equation of motion of a projectile is, y=12x-(3/4)x ^2 The horizontal component of velocity is 3 m/s Given that g=10 m/s^2, what is range of the projectile?
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Oblique projectile equation is
y = xTanθ - gx² / (2u²Cos²θ)
By comparing it with your equation
Tanθ = 12
Then,
Cosθ = 1 / √(145)
Cos²θ = 1/145
Sinθ = 12/√145
3/4 = g / (2u²Cos²θ)
Cos²θ = 4g / (6u²)
1/145 = 4g / (6u²)
u² = 145×4×10 / 6
u² = 966.6
u = 31 m/s
Time of flight = 2uSinθ/g
T = 2×31×(12/√145) / 10
T ≈ 6 s
Range = Horizontal velocity × Time of flight
R = 3m/s × 6s
R = 18 m
y = xTanθ - gx² / (2u²Cos²θ)
By comparing it with your equation
Tanθ = 12
Then,
Cosθ = 1 / √(145)
Cos²θ = 1/145
Sinθ = 12/√145
3/4 = g / (2u²Cos²θ)
Cos²θ = 4g / (6u²)
1/145 = 4g / (6u²)
u² = 145×4×10 / 6
u² = 966.6
u = 31 m/s
Time of flight = 2uSinθ/g
T = 2×31×(12/√145) / 10
T ≈ 6 s
Range = Horizontal velocity × Time of flight
R = 3m/s × 6s
R = 18 m
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