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A projectile is launched with speed of 10m/s at an angle 60 with horizontal from a sloping surface of inclination 30. The range is ?
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A projectile is launched with speed of 10m/s at an angle 60 with horiz...
Projectile Launched from a Sloping Surface
To determine the range of a projectile launched from a sloping surface, we need to consider the initial velocity and angle of projection, as well as the inclination of the surface. In this case, the projectile is launched with a speed of 10 m/s at an angle of 60 degrees with the horizontal from a surface inclined at 30 degrees.
Components of Initial Velocity
The initial velocity of the projectile can be broken down into its horizontal and vertical components. The horizontal component (Vx) is given by Vx = V * cos(θ), where V is the magnitude of the initial velocity and θ is the angle of projection.
In this case, Vx = 10 m/s * cos(60°) = 10 m/s * 0.5 = 5 m/s.
The vertical component (Vy) is given by Vy = V * sin(θ), where V is the magnitude of the initial velocity and θ is the angle of projection.
In this case, Vy = 10 m/s * sin(60°) = 10 m/s * √3/2 ≈ 8.66 m/s.
Effects of Sloping Surface
The inclination of the surface affects the horizontal and vertical motion of the projectile. The vertical acceleration is influenced by the component of gravity acting along the slope, which can be calculated as g * sin(α), where g is the acceleration due to gravity and α is the angle of inclination.
In this case, the vertical acceleration along the slope is 9.8 m/s² * sin(30°) = 4.9 m/s².
The horizontal acceleration is unaffected by the slope, as gravity acts vertically downwards. Therefore, the horizontal acceleration remains zero.
Range Calculation
The range of the projectile can be calculated using the formula R = (Vx * t), where R is the range, Vx is the horizontal component of the initial velocity, and t is the time of flight.
To find the time of flight, we can use the formula t = (2 * Vy) / (g * sin(α)), where Vy is the vertical component of the initial velocity and α is the angle of inclination.
In this case, t = (2 * 8.66 m/s) / (9.8 m/s² * sin(30°)) ≈ 1.77 seconds.
Substituting the values into the range formula, we get R = (5 m/s * 1.77 s) = 8.85 m.
Therefore, the range of the projectile launched from the sloping surface is approximately 8.85 meters.
To determine the range of a projectile launched from a sloping surface, we need to consider the initial velocity and angle of projection, as well as the inclination of the surface. In this case, the projectile is launched with a speed of 10 m/s at an angle of 60 degrees with the horizontal from a surface inclined at 30 degrees.
Components of Initial Velocity
The initial velocity of the projectile can be broken down into its horizontal and vertical components. The horizontal component (Vx) is given by Vx = V * cos(θ), where V is the magnitude of the initial velocity and θ is the angle of projection.
In this case, Vx = 10 m/s * cos(60°) = 10 m/s * 0.5 = 5 m/s.
The vertical component (Vy) is given by Vy = V * sin(θ), where V is the magnitude of the initial velocity and θ is the angle of projection.
In this case, Vy = 10 m/s * sin(60°) = 10 m/s * √3/2 ≈ 8.66 m/s.
Effects of Sloping Surface
The inclination of the surface affects the horizontal and vertical motion of the projectile. The vertical acceleration is influenced by the component of gravity acting along the slope, which can be calculated as g * sin(α), where g is the acceleration due to gravity and α is the angle of inclination.
In this case, the vertical acceleration along the slope is 9.8 m/s² * sin(30°) = 4.9 m/s².
The horizontal acceleration is unaffected by the slope, as gravity acts vertically downwards. Therefore, the horizontal acceleration remains zero.
Range Calculation
The range of the projectile can be calculated using the formula R = (Vx * t), where R is the range, Vx is the horizontal component of the initial velocity, and t is the time of flight.
To find the time of flight, we can use the formula t = (2 * Vy) / (g * sin(α)), where Vy is the vertical component of the initial velocity and α is the angle of inclination.
In this case, t = (2 * 8.66 m/s) / (9.8 m/s² * sin(30°)) ≈ 1.77 seconds.
Substituting the values into the range formula, we get R = (5 m/s * 1.77 s) = 8.85 m.
Therefore, the range of the projectile launched from the sloping surface is approximately 8.85 meters.
Community Answer
A projectile is launched with speed of 10m/s at an angle 60 with horiz...
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A projectile is launched with speed of 10m/s at an angle 60 with horizontal from a sloping surface of inclination 30. The range is ?
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A projectile is launched with speed of 10m/s at an angle 60 with horizontal from a sloping surface of inclination 30. The range is ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A projectile is launched with speed of 10m/s at an angle 60 with horizontal from a sloping surface of inclination 30. The range is ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A projectile is launched with speed of 10m/s at an angle 60 with horizontal from a sloping surface of inclination 30. The range is ?.
A projectile is launched with speed of 10m/s at an angle 60 with horizontal from a sloping surface of inclination 30. The range is ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A projectile is launched with speed of 10m/s at an angle 60 with horizontal from a sloping surface of inclination 30. The range is ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A projectile is launched with speed of 10m/s at an angle 60 with horizontal from a sloping surface of inclination 30. The range is ?.
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