Instantaneous Velocity And Speed

When a ball is thrown vertically upwards, its velocity decreases due to the acceleration due to gravity, which acts in the downward direction. At the highest point of its motion, the velocity of the ball becomes zero because it momentarily stops before reversing direction and starting to fall back down.
Thus, at the highest point:
- The velocity is .
- The acceleration is still due to gravity, which is constant and equal to in the downward direction.
Thus, the correct answer is option (B), 0 and 9.8.

The position vector of the particle is given by: To find the velocity, we differentiate the position vector with respect to time: Differentiating each component: Thus, the velocity vector is: At seconds: The magnitude of the velocity is: Thus, the magnitude of the velocity at seconds is .

If the velocity-time graph cuts the time axis, then at that instant: If it crosses the axis, velocity changes sign: That means the particle reverses its direction of motion.
Hence, the correct answer is:

Concept:
Physics - Kinematics and Derivatives.
Velocity is the first derivative of position with respect to time: .
Step 1: Differentiate the position function.
Given position . $ 3t t=0 t=2 t = 2$ s.

Step 1: Use the formula for escape velocity.
The escape velocity from the surface of a planet is given by the formula: where:
- is the universal gravitational constant,
- is the mass of the planet, and
- is the radius of the planet.

Step 2: Express mass in terms of density.
The mass of a planet is related to its density and volume. The volume of a spherical planet is given by: Thus, the mass of the planet is:
Step 3: Substitute into the formula for escape velocity.
Substitute the expression for mass into the formula for escape velocity: Thus, the escape velocity is proportional to:
Step 4: Find the ratio of escape velocities.
Now, let’s find the ratio of escape velocities for planets A and B. The ratio of escape velocities is: We are given that: - , - . Substituting these values: Thus, the ratio of escape velocities is:

Step 1: Maximum height formula.
For a projectile launched at an angle with initial velocity , the maximum height is given by: For , we have , so:
Step 2: Height at second.
The height at any time is given by: At second, for , we have , so:
Step 3: Ratio of heights.
The required ratio of the maximum height to the height at second is: Simplifying: