Motion In A Straight Line

According to the question,
Velocity of an object half a second before maximum height
= Velocity of an object half a second after maximum height (return journey)

Step 1: Compute velocity of center of mass
The velocity of the center of mass is given by: The x-component is: The y-component is: Step 2: Compute magnitude of
Thus, the correct answer is m/s.

Step 1: Using Free Fall Equation

For the first pebble: sec
Step 3: Conclusion

Since the height is 20 m, Option (1) is correct.

Step 1: Use SHM potential energy formula
Potential energy for SHM is: Given , comparing with the standard equation, we get: Step 2: Compute using angular frequency
The angular frequency is: Squaring both sides: Substituting kg and rad/s, Step 3: Compute
Thus, the correct answer is J/m .

Step 1: Understanding the Concept:
This problem involves motion with non-uniform acceleration. The motion must be broken down into two distinct phases: the first phase with constant positive acceleration and the second phase with constant negative acceleration. We will use the standard kinematic equations.
Step 2: Phase 1 (from t = 0 to t = 3 s):
The particle starts from rest ( ) with acceleration ms⁻².
First, we find the velocity ( ) and displacement ( ) at the end of this phase (at s).
So, at t = 3 s, the particle is at a position of 9 m from the start and is moving with a velocity of 6 m/s.
Step 3: Phase 2 (for t>3 s):
Now, the acceleration is reversed, so ms⁻². The initial velocity for this phase is the final velocity from Phase 1, i.e., m/s. The particle is currently at m. It needs to return to its initial position, which means its displacement in this second phase must be m.
Let be the time taken for this second phase. We use the displacement equation:
Rearranging this into a quadratic equation:
We solve for using the quadratic formula :
Since time cannot be negative, we take the positive root:
Step 4: Final Answer:
The total time from the beginning of the motion is the sum of the times for both phases:
Thus, the correct option is (C).

The motion consists of two parts.
Part 1: Freefall for a distance of m.
We need to find the time taken ( ) and the final velocity ( ) for this part. Using the kinematic equation , with initial velocity .
.
m/s. This is the uniform velocity for the second part of the journey.
To find the time , use .
s.
Part 2: Uniform motion for the remaining distance.
Total height km = 2000 m.
Remaining height m.
The person travels this distance with a uniform velocity m/s.
The time taken for this part ( ) is given by distance/speed.
s.
Total time taken to reach the ground is the sum of the times for both parts.
Total time s.

Step 1: Use the formula for displacement in the second.}
The displacement, , of a particle undergoing uniformly accelerated motion during the second is given by: where is the initial velocity and is the constant acceleration.

Step 2: Set up equations using the given data.
For the third second ( ), the displacement is 10 m. For the fifth second ( ), the displacement is 18 m.

Step 3: Solve the system of equations for and .
Subtract Equation (1) from Equation (2): Substitute the value of back into Equation (1):

Step 4: Calculate the speed at t=4s.
The speed (velocity) at time is given by the kinematic equation . We need to find the speed at s.