Given that \(y = cx - a{x^2}\) The angle of projection is \(\tan \theta = \frac{{dy}}{{dx}}\left( {x = 0} \right)\) \( \Rightarrow \frac{{dy}}{{dx}} = c - 2ax = c\) \( \Rightarrow \theta = {\tan ^{ - 1}}\left( c \right)\)
PHXI04:MOTION IN A PLANE
361962
The velocity of a projectile at the initial point \(A\) is \((2\hat i + 3\hat j)\,m{\rm{/}}s\). Its velocity (in \(m{\rm{/}}s\) ) at point \(B\) is
1 \(-2 \hat{i}+3 \hat{j}\)
2 \(2 \hat{i}-3 \hat{j}\)
3 \(2 \hat{i}+3 \hat{j}\)
4 \(-2 \hat{i}-3 \hat{j}\)
Explanation:
At a point \(A\), initial velocity \(V_{A}=2 \hat{i}+3 \hat{j}\) At a point \(B\), final velocity \(V_{B}=2 \hat{i}-3 \hat{j}\) \(\therefore\) The \(X\)-component of velocity remains constant and the \(Y\)-component reverse in direction of \(\left( { - \hat j} \right)\).
MHTCET - 2021
PHXI04:MOTION IN A PLANE
361963
What determines the nature of the path followed by the particle?
1 Speed
2 Velocity
3 Acceleration
4 Both (2) and (3)
Explanation:
The nature of path is decided by velocity and acceleration.Velocity is ratio of displacement to time. It is a vector quantity possessing both magnitude and direction. When velocity of a particle changes, the particle is said to undergo an acceleration. The instantaneous acceleration is the derivative of the velocity with respect to time. Acceleration is also a vector quantity, possessing both magnitude and direction. Depending on variation of velocity and acceleration, the path could be a straight line, circle or a parabola.
PHXI04:MOTION IN A PLANE
361964
A particle is projected with velocity \(u\) at angle \(\theta \) with horizontal at \(t = 0\). What is the magnitude of change in the velocity of the particle when it is at maximum height?
1 \(\frac{{u\cos \theta }}{2}\)
2 \(u\cos \theta \)
3 \(u\sin \theta \)
4 \({\rm{None}}\,\,{\rm{of}}\,\,{\rm{these}}\)
Explanation:
\({\overrightarrow u _i} = u\cos \theta \hat i + u\sin \theta \hat j\) \({\overrightarrow u _f} = u\cos \theta \hat i\) \(\Delta \overrightarrow u = {\overrightarrow u _f} = - u\sin \theta \hat j\) \(\left| {\Delta \overrightarrow u } \right| = u\sin \theta \)
Given that \(y = cx - a{x^2}\) The angle of projection is \(\tan \theta = \frac{{dy}}{{dx}}\left( {x = 0} \right)\) \( \Rightarrow \frac{{dy}}{{dx}} = c - 2ax = c\) \( \Rightarrow \theta = {\tan ^{ - 1}}\left( c \right)\)
PHXI04:MOTION IN A PLANE
361962
The velocity of a projectile at the initial point \(A\) is \((2\hat i + 3\hat j)\,m{\rm{/}}s\). Its velocity (in \(m{\rm{/}}s\) ) at point \(B\) is
1 \(-2 \hat{i}+3 \hat{j}\)
2 \(2 \hat{i}-3 \hat{j}\)
3 \(2 \hat{i}+3 \hat{j}\)
4 \(-2 \hat{i}-3 \hat{j}\)
Explanation:
At a point \(A\), initial velocity \(V_{A}=2 \hat{i}+3 \hat{j}\) At a point \(B\), final velocity \(V_{B}=2 \hat{i}-3 \hat{j}\) \(\therefore\) The \(X\)-component of velocity remains constant and the \(Y\)-component reverse in direction of \(\left( { - \hat j} \right)\).
MHTCET - 2021
PHXI04:MOTION IN A PLANE
361963
What determines the nature of the path followed by the particle?
1 Speed
2 Velocity
3 Acceleration
4 Both (2) and (3)
Explanation:
The nature of path is decided by velocity and acceleration.Velocity is ratio of displacement to time. It is a vector quantity possessing both magnitude and direction. When velocity of a particle changes, the particle is said to undergo an acceleration. The instantaneous acceleration is the derivative of the velocity with respect to time. Acceleration is also a vector quantity, possessing both magnitude and direction. Depending on variation of velocity and acceleration, the path could be a straight line, circle or a parabola.
PHXI04:MOTION IN A PLANE
361964
A particle is projected with velocity \(u\) at angle \(\theta \) with horizontal at \(t = 0\). What is the magnitude of change in the velocity of the particle when it is at maximum height?
1 \(\frac{{u\cos \theta }}{2}\)
2 \(u\cos \theta \)
3 \(u\sin \theta \)
4 \({\rm{None}}\,\,{\rm{of}}\,\,{\rm{these}}\)
Explanation:
\({\overrightarrow u _i} = u\cos \theta \hat i + u\sin \theta \hat j\) \({\overrightarrow u _f} = u\cos \theta \hat i\) \(\Delta \overrightarrow u = {\overrightarrow u _f} = - u\sin \theta \hat j\) \(\left| {\Delta \overrightarrow u } \right| = u\sin \theta \)
Given that \(y = cx - a{x^2}\) The angle of projection is \(\tan \theta = \frac{{dy}}{{dx}}\left( {x = 0} \right)\) \( \Rightarrow \frac{{dy}}{{dx}} = c - 2ax = c\) \( \Rightarrow \theta = {\tan ^{ - 1}}\left( c \right)\)
PHXI04:MOTION IN A PLANE
361962
The velocity of a projectile at the initial point \(A\) is \((2\hat i + 3\hat j)\,m{\rm{/}}s\). Its velocity (in \(m{\rm{/}}s\) ) at point \(B\) is
1 \(-2 \hat{i}+3 \hat{j}\)
2 \(2 \hat{i}-3 \hat{j}\)
3 \(2 \hat{i}+3 \hat{j}\)
4 \(-2 \hat{i}-3 \hat{j}\)
Explanation:
At a point \(A\), initial velocity \(V_{A}=2 \hat{i}+3 \hat{j}\) At a point \(B\), final velocity \(V_{B}=2 \hat{i}-3 \hat{j}\) \(\therefore\) The \(X\)-component of velocity remains constant and the \(Y\)-component reverse in direction of \(\left( { - \hat j} \right)\).
MHTCET - 2021
PHXI04:MOTION IN A PLANE
361963
What determines the nature of the path followed by the particle?
1 Speed
2 Velocity
3 Acceleration
4 Both (2) and (3)
Explanation:
The nature of path is decided by velocity and acceleration.Velocity is ratio of displacement to time. It is a vector quantity possessing both magnitude and direction. When velocity of a particle changes, the particle is said to undergo an acceleration. The instantaneous acceleration is the derivative of the velocity with respect to time. Acceleration is also a vector quantity, possessing both magnitude and direction. Depending on variation of velocity and acceleration, the path could be a straight line, circle or a parabola.
PHXI04:MOTION IN A PLANE
361964
A particle is projected with velocity \(u\) at angle \(\theta \) with horizontal at \(t = 0\). What is the magnitude of change in the velocity of the particle when it is at maximum height?
1 \(\frac{{u\cos \theta }}{2}\)
2 \(u\cos \theta \)
3 \(u\sin \theta \)
4 \({\rm{None}}\,\,{\rm{of}}\,\,{\rm{these}}\)
Explanation:
\({\overrightarrow u _i} = u\cos \theta \hat i + u\sin \theta \hat j\) \({\overrightarrow u _f} = u\cos \theta \hat i\) \(\Delta \overrightarrow u = {\overrightarrow u _f} = - u\sin \theta \hat j\) \(\left| {\Delta \overrightarrow u } \right| = u\sin \theta \)
Given that \(y = cx - a{x^2}\) The angle of projection is \(\tan \theta = \frac{{dy}}{{dx}}\left( {x = 0} \right)\) \( \Rightarrow \frac{{dy}}{{dx}} = c - 2ax = c\) \( \Rightarrow \theta = {\tan ^{ - 1}}\left( c \right)\)
PHXI04:MOTION IN A PLANE
361962
The velocity of a projectile at the initial point \(A\) is \((2\hat i + 3\hat j)\,m{\rm{/}}s\). Its velocity (in \(m{\rm{/}}s\) ) at point \(B\) is
1 \(-2 \hat{i}+3 \hat{j}\)
2 \(2 \hat{i}-3 \hat{j}\)
3 \(2 \hat{i}+3 \hat{j}\)
4 \(-2 \hat{i}-3 \hat{j}\)
Explanation:
At a point \(A\), initial velocity \(V_{A}=2 \hat{i}+3 \hat{j}\) At a point \(B\), final velocity \(V_{B}=2 \hat{i}-3 \hat{j}\) \(\therefore\) The \(X\)-component of velocity remains constant and the \(Y\)-component reverse in direction of \(\left( { - \hat j} \right)\).
MHTCET - 2021
PHXI04:MOTION IN A PLANE
361963
What determines the nature of the path followed by the particle?
1 Speed
2 Velocity
3 Acceleration
4 Both (2) and (3)
Explanation:
The nature of path is decided by velocity and acceleration.Velocity is ratio of displacement to time. It is a vector quantity possessing both magnitude and direction. When velocity of a particle changes, the particle is said to undergo an acceleration. The instantaneous acceleration is the derivative of the velocity with respect to time. Acceleration is also a vector quantity, possessing both magnitude and direction. Depending on variation of velocity and acceleration, the path could be a straight line, circle or a parabola.
PHXI04:MOTION IN A PLANE
361964
A particle is projected with velocity \(u\) at angle \(\theta \) with horizontal at \(t = 0\). What is the magnitude of change in the velocity of the particle when it is at maximum height?
1 \(\frac{{u\cos \theta }}{2}\)
2 \(u\cos \theta \)
3 \(u\sin \theta \)
4 \({\rm{None}}\,\,{\rm{of}}\,\,{\rm{these}}\)
Explanation:
\({\overrightarrow u _i} = u\cos \theta \hat i + u\sin \theta \hat j\) \({\overrightarrow u _f} = u\cos \theta \hat i\) \(\Delta \overrightarrow u = {\overrightarrow u _f} = - u\sin \theta \hat j\) \(\left| {\Delta \overrightarrow u } \right| = u\sin \theta \)
