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A projectile is fired at an angle of 45 degree with the horizontal .Elevation angle of the projectile at its highest point as seen from the point of projection is: i) 45 degree ii) 60 deg iii) tan ^-1(0.25) iv) tan^-1(0.85) Can you help me pls .?
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A projectile is fired at an angle of 45 degree with the horizontal .El...
**Projectile Motion**
When a projectile is fired at an angle with the horizontal, it follows a curved trajectory known as projectile motion. The motion can be divided into horizontal and vertical components.
**Horizontal Component of Motion**
The horizontal component of motion remains constant throughout the flight of the projectile. It is given by the equation:
\(V_x = V_0 \cos(\theta)\)
Where:
\(V_x\) is the horizontal component of velocity,
\(V_0\) is the initial velocity of the projectile, and
\(\theta\) is the angle of projection.
**Vertical Component of Motion**
The vertical component of motion is influenced by gravity and changes over time. It can be analyzed using the equations of motion:
\(V_y = V_0 \sin(\theta) - gt\)
\(y = V_0 \sin(\theta) t - \frac{1}{2} g t^2\)
Where:
\(V_y\) is the vertical component of velocity,
\(g\) is the acceleration due to gravity (approximately 9.8 m/s²),
\(t\) is the time, and
\(y\) is the vertical displacement.
**Elevation Angle at Highest Point**
At the highest point of the projectile's trajectory, the vertical component of velocity becomes zero. This occurs when \(V_y = 0\):
\(V_0 \sin(\theta) - gt = 0\)
Solving for \(t\), we get:
\(t = \frac{V_0 \sin(\theta)}{g}\)
Substituting this value of \(t\) into the equation for \(y\), we can find the vertical displacement at the highest point:
\(y = V_0 \sin(\theta) \left(\frac{V_0 \sin(\theta)}{g}\right) - \frac{1}{2} g \left(\frac{V_0 \sin(\theta)}{g}\right)^2\)
\(y = \frac{V_0^2 \sin^2(\theta)}{2g}\)
To find the elevation angle at the highest point as seen from the point of projection, we need to calculate the angle whose tangent is equal to the ratio of the vertical displacement to the horizontal displacement.
\(\tan(\text{elevation angle}) = \frac{y}{x}\)
Considering the projectile's initial velocity is the same in both horizontal and vertical directions, the horizontal displacement (\(x\)) can be expressed in terms of the initial velocity and time:
\(x = V_0 \cos(\theta) \times \frac{2V_0 \sin(\theta)}{g}\)
Simplifying, we get:
\(x = \frac{2V_0^2 \sin(\theta) \cos(\theta)}{g}\)
Substituting the values of \(x\) and \(y\) in the equation for the tangent of the elevation angle:
\(\tan(\text{elevation angle}) = \frac{\frac{V_0^2 \sin^2(\theta)}{2g}}{\frac{2V_0^2 \sin(\theta) \cos(\theta)}{g}}\)
\(\tan(\text{elevation angle}) = \frac{\sin(\theta)}{2\cos(\theta)}\)
\(\tan(\text{elevation angle})
When a projectile is fired at an angle with the horizontal, it follows a curved trajectory known as projectile motion. The motion can be divided into horizontal and vertical components.
**Horizontal Component of Motion**
The horizontal component of motion remains constant throughout the flight of the projectile. It is given by the equation:
\(V_x = V_0 \cos(\theta)\)
Where:
\(V_x\) is the horizontal component of velocity,
\(V_0\) is the initial velocity of the projectile, and
\(\theta\) is the angle of projection.
**Vertical Component of Motion**
The vertical component of motion is influenced by gravity and changes over time. It can be analyzed using the equations of motion:
\(V_y = V_0 \sin(\theta) - gt\)
\(y = V_0 \sin(\theta) t - \frac{1}{2} g t^2\)
Where:
\(V_y\) is the vertical component of velocity,
\(g\) is the acceleration due to gravity (approximately 9.8 m/s²),
\(t\) is the time, and
\(y\) is the vertical displacement.
**Elevation Angle at Highest Point**
At the highest point of the projectile's trajectory, the vertical component of velocity becomes zero. This occurs when \(V_y = 0\):
\(V_0 \sin(\theta) - gt = 0\)
Solving for \(t\), we get:
\(t = \frac{V_0 \sin(\theta)}{g}\)
Substituting this value of \(t\) into the equation for \(y\), we can find the vertical displacement at the highest point:
\(y = V_0 \sin(\theta) \left(\frac{V_0 \sin(\theta)}{g}\right) - \frac{1}{2} g \left(\frac{V_0 \sin(\theta)}{g}\right)^2\)
\(y = \frac{V_0^2 \sin^2(\theta)}{2g}\)
To find the elevation angle at the highest point as seen from the point of projection, we need to calculate the angle whose tangent is equal to the ratio of the vertical displacement to the horizontal displacement.
\(\tan(\text{elevation angle}) = \frac{y}{x}\)
Considering the projectile's initial velocity is the same in both horizontal and vertical directions, the horizontal displacement (\(x\)) can be expressed in terms of the initial velocity and time:
\(x = V_0 \cos(\theta) \times \frac{2V_0 \sin(\theta)}{g}\)
Simplifying, we get:
\(x = \frac{2V_0^2 \sin(\theta) \cos(\theta)}{g}\)
Substituting the values of \(x\) and \(y\) in the equation for the tangent of the elevation angle:
\(\tan(\text{elevation angle}) = \frac{\frac{V_0^2 \sin^2(\theta)}{2g}}{\frac{2V_0^2 \sin(\theta) \cos(\theta)}{g}}\)
\(\tan(\text{elevation angle}) = \frac{\sin(\theta)}{2\cos(\theta)}\)
\(\tan(\text{elevation angle})
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A projectile is fired at an angle of 45 degree with the horizontal .Elevation angle of the projectile at its highest point as seen from the point of projection is: i) 45 degree ii) 60 deg iii) tan ^-1(0.25) iv) tan^-1(0.85) Can you help me pls .?
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A projectile is fired at an angle of 45 degree with the horizontal .Elevation angle of the projectile at its highest point as seen from the point of projection is: i) 45 degree ii) 60 deg iii) tan ^-1(0.25) iv) tan^-1(0.85) Can you help me pls .? 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 fired at an angle of 45 degree with the horizontal .Elevation angle of the projectile at its highest point as seen from the point of projection is: i) 45 degree ii) 60 deg iii) tan ^-1(0.25) iv) tan^-1(0.85) Can you help me pls .? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A projectile is fired at an angle of 45 degree with the horizontal .Elevation angle of the projectile at its highest point as seen from the point of projection is: i) 45 degree ii) 60 deg iii) tan ^-1(0.25) iv) tan^-1(0.85) Can you help me pls .?.
A projectile is fired at an angle of 45 degree with the horizontal .Elevation angle of the projectile at its highest point as seen from the point of projection is: i) 45 degree ii) 60 deg iii) tan ^-1(0.25) iv) tan^-1(0.85) Can you help me pls .? 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 fired at an angle of 45 degree with the horizontal .Elevation angle of the projectile at its highest point as seen from the point of projection is: i) 45 degree ii) 60 deg iii) tan ^-1(0.25) iv) tan^-1(0.85) Can you help me pls .? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A projectile is fired at an angle of 45 degree with the horizontal .Elevation angle of the projectile at its highest point as seen from the point of projection is: i) 45 degree ii) 60 deg iii) tan ^-1(0.25) iv) tan^-1(0.85) Can you help me pls .?.
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