For a body projected at an angle with the horizontal from the ground, choose the correct statement.
Show Hint
In projectile motion, analyze the components of velocity and the energies (potential and kinetic) at different points of the trajectory, especially at the highest point.
- The Kinetic Energy (K.E.) is zero at the highest point of projectile motion.
- The horizontal component of velocity is zero at the highest point.
- The vertical component of momentum is maximum at the highest point.
- Gravitational potential energy is maximum at the highest point.
The Correct Option is D
Approach Solution - 1
Step 1: Analyze the Projectile Motion at the Highest Point
When a body is projected at an angle with the horizontal, it follows a parabolic trajectory. At the highest point of its trajectory:
- The vertical component of velocity (\(v_y\)) is zero.
- The horizontal component of velocity (\(v_x\)) remains constant throughout the motion and is equal to \(u \cos \theta\), where \(u\) is the initial velocity and \(\theta\) is the angle of projection.
Step 2: Analyze the Gravitational Potential Energy
The gravitational potential energy (\(U\)) of a body at a height \(h\) above the ground is given by:
\[ U = mgh \]
where \(m\) is the mass of the body and \(g\) is the acceleration due to gravity. Since \(h\) is maximum at the highest point, the gravitational potential energy is also maximum at the highest point.
Step 3: Analyze the Horizontal Component of Velocity
The horizontal component of velocity remains constant throughout the projectile motion. It does not become zero at the highest point.
Step 4: Analyze the Vertical Component of Momentum
The vertical component of momentum is given by \(mv_y\). Since \(v_y = 0\) at the highest point, the vertical component of momentum is also zero at the highest point, not maximum.
Step 5: Analyze the Kinetic Energy
The kinetic energy (\(KE\)) of a body is given by:
\[ KE = \frac{1}{2} mv^2 \]
At the highest point, \(v_y = 0\), but \(v_x = u \cos \theta \neq 0\). Therefore, the total velocity \(v\) is not zero, and hence \(KE\) is not zero. The \(KE\) at the highest point is minimum but not zero.
Conclusion: Only statement (4) is correct. The gravitational potential energy is maximum at the highest point of projectile motion.
Approach Solution -2
At highest point
\(Vy=0\)
\(Vx=ux=ucosθ\)
\(Ug=mgh\), it is maximum at \(H_{max}\)
The Correct Option is (D): Gravitational potential energy is maximum at the highest point.
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration