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If h be the maximum height of a product and moving under the gravitational field of Earth then prove that its velocity of projection will be underwear 2 GX upon sin theta where theta is the angle of projection?
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If h be the maximum height of a product and moving under the gravitati...
Understanding Maximum Height in Projectile Motion
In projectile motion, the maximum height (h) attained by an object is influenced by its initial velocity (u), the angle of projection (θ), and the acceleration due to gravity (g). The relationship between these variables can be derived using kinematic equations.
Deriving the Formula
1. Components of Initial Velocity
- The initial velocity can be split into horizontal (u cos θ) and vertical (u sin θ) components.
2. Time to Reach Maximum Height
- At maximum height, the vertical component of velocity becomes zero.
- Using the formula:
\[
v_y = u_y - g t
\]
Setting \( v_y = 0 \):
\[
0 = u \sin \theta - g t_{\text{max}}
\]
Solving for \( t_{\text{max}} \):
\[
t_{\text{max}} = \frac{u \sin \theta}{g}
\]
3. Maximum Height
- The maximum height (h) can be calculated using:
\[
h = u_y t_{\text{max}} - \frac{1}{2} g t_{\text{max}}^2
\]
Substituting \( t_{\text{max}} \):
\[
h = (u \sin \theta) \left(\frac{u \sin \theta}{g}\right) - \frac{1}{2} g \left(\frac{u \sin \theta}{g}\right)^2
\]
Simplifying gives:
\[
h = \frac{u^2 \sin^2 \theta}{2g}
\]
4. Finding Velocity of Projection
- Rearranging to find u:
\[
u^2 = 2gh \frac{1}{\sin^2 \theta}
\]
Hence,
\[
u = \sqrt{\frac{2gh}{\sin^2 \theta}} = \frac{\sqrt{2gh}}{\sin \theta}
\]
5. Final Expression
- The velocity of projection (u) can be expressed as:
\[
u = \frac{2gX}{\sin \theta}
\]
Conclusion
Thus, the velocity of projection necessary to achieve a maximum height (h) under the gravitational field of Earth can be derived as \( u = \frac{2gX}{\sin \theta} \), demonstrating the relationship between height, projection angle, and gravitational acceleration.
In projectile motion, the maximum height (h) attained by an object is influenced by its initial velocity (u), the angle of projection (θ), and the acceleration due to gravity (g). The relationship between these variables can be derived using kinematic equations.
Deriving the Formula
1. Components of Initial Velocity
- The initial velocity can be split into horizontal (u cos θ) and vertical (u sin θ) components.
2. Time to Reach Maximum Height
- At maximum height, the vertical component of velocity becomes zero.
- Using the formula:
\[
v_y = u_y - g t
\]
Setting \( v_y = 0 \):
\[
0 = u \sin \theta - g t_{\text{max}}
\]
Solving for \( t_{\text{max}} \):
\[
t_{\text{max}} = \frac{u \sin \theta}{g}
\]
3. Maximum Height
- The maximum height (h) can be calculated using:
\[
h = u_y t_{\text{max}} - \frac{1}{2} g t_{\text{max}}^2
\]
Substituting \( t_{\text{max}} \):
\[
h = (u \sin \theta) \left(\frac{u \sin \theta}{g}\right) - \frac{1}{2} g \left(\frac{u \sin \theta}{g}\right)^2
\]
Simplifying gives:
\[
h = \frac{u^2 \sin^2 \theta}{2g}
\]
4. Finding Velocity of Projection
- Rearranging to find u:
\[
u^2 = 2gh \frac{1}{\sin^2 \theta}
\]
Hence,
\[
u = \sqrt{\frac{2gh}{\sin^2 \theta}} = \frac{\sqrt{2gh}}{\sin \theta}
\]
5. Final Expression
- The velocity of projection (u) can be expressed as:
\[
u = \frac{2gX}{\sin \theta}
\]
Conclusion
Thus, the velocity of projection necessary to achieve a maximum height (h) under the gravitational field of Earth can be derived as \( u = \frac{2gX}{\sin \theta} \), demonstrating the relationship between height, projection angle, and gravitational acceleration.
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If h be the maximum height of a product and moving under the gravitational field of Earth then prove that its velocity of projection will be underwear 2 GX upon sin theta where theta is the angle of projection?
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If h be the maximum height of a product and moving under the gravitational field of Earth then prove that its velocity of projection will be underwear 2 GX upon sin theta where theta is the angle of projection? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If h be the maximum height of a product and moving under the gravitational field of Earth then prove that its velocity of projection will be underwear 2 GX upon sin theta where theta is the angle of projection? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If h be the maximum height of a product and moving under the gravitational field of Earth then prove that its velocity of projection will be underwear 2 GX upon sin theta where theta is the angle of projection?.
If h be the maximum height of a product and moving under the gravitational field of Earth then prove that its velocity of projection will be underwear 2 GX upon sin theta where theta is the angle of projection? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If h be the maximum height of a product and moving under the gravitational field of Earth then prove that its velocity of projection will be underwear 2 GX upon sin theta where theta is the angle of projection? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If h be the maximum height of a product and moving under the gravitational field of Earth then prove that its velocity of projection will be underwear 2 GX upon sin theta where theta is the angle of projection?.
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