362009
The acceleration of one projectile relative to another projectile is
1 \( - g\)
2 \(g\)
3 \(2g\)
4 \({\mathop{\rm zero}\nolimits} \)
Explanation:
PHXI04:MOTION IN A PLANE
362010
A missile is fired for maximum range with an initial velocity of \(20\,m/s.\) If \(g = 10\,m/{s^2}\), the range of the missile is
1 \(40\,m\)
2 \(20\,m\)
3 \(60\,m\)
4 \(50\,m\)
Explanation:
For maximum range of projectile, \(\theta \) will be \(45^\circ \) by the law of projectile motion. So, maximum range, \({R_{\max }} = \frac{{{u^2}}}{g}\) Given, \(u = 20\,m{s^{ - 1}}\) and \(g = 10\,m{s^{ - 2}}\) \({R_{\max }} = \frac{{{{(20)}^2}}}{{10}} = \frac{{400}}{{10}}\) \({R_{\max }} = 40\,m\)
PHXI04:MOTION IN A PLANE
362011
Assertion : When the velocity of projection of a body is made \(n\) times, its time of flight becomes \(n\) times. Reason : Range of projectile does not depend on the initial velocity of a body.
1 Both assertion and reason are correct and reason isthe correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Time of flight \(T = \frac{{2u\sin \theta }}{g}\). When \(u\) is made \(n\) times then \(T\) also become \(n\) times. Option (3) is correct.
PHXI04:MOTION IN A PLANE
362012
The horizontal range of a projectile is \(4\sqrt 3 \) times its maximum height. Its angle of projection will be
362009
The acceleration of one projectile relative to another projectile is
1 \( - g\)
2 \(g\)
3 \(2g\)
4 \({\mathop{\rm zero}\nolimits} \)
Explanation:
PHXI04:MOTION IN A PLANE
362010
A missile is fired for maximum range with an initial velocity of \(20\,m/s.\) If \(g = 10\,m/{s^2}\), the range of the missile is
1 \(40\,m\)
2 \(20\,m\)
3 \(60\,m\)
4 \(50\,m\)
Explanation:
For maximum range of projectile, \(\theta \) will be \(45^\circ \) by the law of projectile motion. So, maximum range, \({R_{\max }} = \frac{{{u^2}}}{g}\) Given, \(u = 20\,m{s^{ - 1}}\) and \(g = 10\,m{s^{ - 2}}\) \({R_{\max }} = \frac{{{{(20)}^2}}}{{10}} = \frac{{400}}{{10}}\) \({R_{\max }} = 40\,m\)
PHXI04:MOTION IN A PLANE
362011
Assertion : When the velocity of projection of a body is made \(n\) times, its time of flight becomes \(n\) times. Reason : Range of projectile does not depend on the initial velocity of a body.
1 Both assertion and reason are correct and reason isthe correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Time of flight \(T = \frac{{2u\sin \theta }}{g}\). When \(u\) is made \(n\) times then \(T\) also become \(n\) times. Option (3) is correct.
PHXI04:MOTION IN A PLANE
362012
The horizontal range of a projectile is \(4\sqrt 3 \) times its maximum height. Its angle of projection will be
362009
The acceleration of one projectile relative to another projectile is
1 \( - g\)
2 \(g\)
3 \(2g\)
4 \({\mathop{\rm zero}\nolimits} \)
Explanation:
PHXI04:MOTION IN A PLANE
362010
A missile is fired for maximum range with an initial velocity of \(20\,m/s.\) If \(g = 10\,m/{s^2}\), the range of the missile is
1 \(40\,m\)
2 \(20\,m\)
3 \(60\,m\)
4 \(50\,m\)
Explanation:
For maximum range of projectile, \(\theta \) will be \(45^\circ \) by the law of projectile motion. So, maximum range, \({R_{\max }} = \frac{{{u^2}}}{g}\) Given, \(u = 20\,m{s^{ - 1}}\) and \(g = 10\,m{s^{ - 2}}\) \({R_{\max }} = \frac{{{{(20)}^2}}}{{10}} = \frac{{400}}{{10}}\) \({R_{\max }} = 40\,m\)
PHXI04:MOTION IN A PLANE
362011
Assertion : When the velocity of projection of a body is made \(n\) times, its time of flight becomes \(n\) times. Reason : Range of projectile does not depend on the initial velocity of a body.
1 Both assertion and reason are correct and reason isthe correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Time of flight \(T = \frac{{2u\sin \theta }}{g}\). When \(u\) is made \(n\) times then \(T\) also become \(n\) times. Option (3) is correct.
PHXI04:MOTION IN A PLANE
362012
The horizontal range of a projectile is \(4\sqrt 3 \) times its maximum height. Its angle of projection will be
362009
The acceleration of one projectile relative to another projectile is
1 \( - g\)
2 \(g\)
3 \(2g\)
4 \({\mathop{\rm zero}\nolimits} \)
Explanation:
PHXI04:MOTION IN A PLANE
362010
A missile is fired for maximum range with an initial velocity of \(20\,m/s.\) If \(g = 10\,m/{s^2}\), the range of the missile is
1 \(40\,m\)
2 \(20\,m\)
3 \(60\,m\)
4 \(50\,m\)
Explanation:
For maximum range of projectile, \(\theta \) will be \(45^\circ \) by the law of projectile motion. So, maximum range, \({R_{\max }} = \frac{{{u^2}}}{g}\) Given, \(u = 20\,m{s^{ - 1}}\) and \(g = 10\,m{s^{ - 2}}\) \({R_{\max }} = \frac{{{{(20)}^2}}}{{10}} = \frac{{400}}{{10}}\) \({R_{\max }} = 40\,m\)
PHXI04:MOTION IN A PLANE
362011
Assertion : When the velocity of projection of a body is made \(n\) times, its time of flight becomes \(n\) times. Reason : Range of projectile does not depend on the initial velocity of a body.
1 Both assertion and reason are correct and reason isthe correct explanation of assertion.
2 Both assertion and reason are correct but reason is not the correct explanation of assertion.
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Time of flight \(T = \frac{{2u\sin \theta }}{g}\). When \(u\) is made \(n\) times then \(T\) also become \(n\) times. Option (3) is correct.
PHXI04:MOTION IN A PLANE
362012
The horizontal range of a projectile is \(4\sqrt 3 \) times its maximum height. Its angle of projection will be
