The following plots show variation of velocity (v), with time (t), of a ball thrown vertically upward, and falling back. Which of the following plots is/are correct?
Show Hint
A velocity-time graph for any object under constant acceleration must be a straight line. If the graph "bounces" back to the positive side (like plot B), it represents a Speed-time graph, not a Velocity-time graph.
Step 1: Understanding the Concept:
When a ball is thrown vertically upward, it is subject to a constant acceleration due to gravity ($g$) acting downwards. Velocity is a vector quantity, meaning its direction matters. Step 2: Key Formula or Approach:
Using the first equation of motion:
\[ v = u + at \]
Since $a = -g$ (taking upward as positive):
\[ v = u - gt \]
This is a linear equation of the form $y = mx + c$, representing a straight line with a negative slope. Step 3: Detailed Explanation:
1. At $t = 0$: The ball has an initial positive velocity ($+u$).
2. Moving Upward: The velocity decreases linearly until it becomes zero at the highest point.
3. At the Peak: $v = 0$.
4. Moving Downward: The velocity becomes negative and increases in magnitude (speeding up in the downward direction).
5. The graph must be a single straight line crossing from the positive quadrant to the negative quadrant. Only Plot C correctly shows this linear transition with a constant negative slope. Step 4: Final Answer:
The correct plot is C only.
