If an object is thrown vertically upwards, what will be its velocity when it reaches its maximum height, assuming motion under gravity and neglecting air resistance?

Introduction / Context:
Throwing an object vertically upwards is a standard motion studied in kinematics. As the object rises, gravity acts downward, slowing it until it momentarily comes to rest at the highest point before falling back down. Understanding what happens to the velocity at the maximum height is essential for solving many problems involving upward and downward motion under gravity.

Given Data / Assumptions:

• The object is projected vertically upward with some initial velocity.
• The only significant force acting on the object during its flight is the gravitational force.
• Air resistance is neglected, so acceleration due to gravity is taken as constant and downward.
• We are asked for the velocity exactly at the maximum height of the trajectory.

Concept / Approach:
As the object moves upward, gravity produces a constant downward acceleration, which continuously reduces the upward velocity. At the maximum height, the object changes its direction of motion from upward to downward. At the exact turning point, the instantaneous velocity must be zero, otherwise the object would still be moving upwards or downwards. In equations of motion, this is seen when using v^2 = u^2 + 2 * a * s; at the highest point, we can set v = 0 to find the maximum height. Therefore, the magnitude of velocity at the maximum height is zero.

Step-by-Step Solution:
Step 1: Consider the motion of the object as it travels upward against gravity.Step 2: Gravity provides a constant downward acceleration g, which reduces the upward velocity over time.Step 3: As long as the object is moving upward, its velocity is positive (upward) but decreasing in magnitude.Step 4: At the maximum height, the object stops rising and is about to start falling back down.Step 5: At this turning point, the instantaneous velocity must be zero; otherwise, motion would continue upward or downward instead of changing direction.Step 6: Therefore, the velocity at the maximum height is 0 m/s.

Verification / Alternative check:
Using the kinematic equation v = u + a * t, with upward taken as positive and acceleration due to gravity as a = -g, we can describe the upward journey. At some time t_max, the object reaches maximum height, where v = 0. Setting v = 0 gives 0 = u - g * t_max, so t_max = u / g. The equation shows that the final velocity at the highest point is exactly zero, independent of the specific values of u or g. Similarly, using v^2 = u^2 + 2 * a * s and setting v = 0 allows calculation of maximum height, again confirming that v is zero at that point.

Why Other Options Are Wrong:
The values 4.9 m/s, 14.7 m/s and 20 m/s all represent non zero speeds. Choosing any of these would imply that the object is still moving when it is supposedly at the maximum height, which is not correct for the idealised case of vertical motion under gravity. These numbers may remind students of values like g * t, but they do not apply at the turning point where the direction reverses.

Common Pitfalls:
Some students confuse the concept of speed with acceleration and think that the object still has some upward speed at the top because gravity is still acting. Gravity indeed acts continuously, but at the highest point, the velocity is momentarily zero while acceleration is still downward. Others may misinterpret diagrams and think the object stops only when it comes back to the starting point. To avoid these mistakes, always remember that at the turnaround point in vertical motion, the instantaneous velocity is zero.

Final Answer:
At its maximum height, the velocity of the object is 0 m/s.