A body of mass 20g projected vertically upwards in vacuum returns to the point of projection after 1.2s. [g =…

Analyze the question: A 20 g body is thrown straight up and returns to its starting point after 1.2 s. We need its potential energy at the highest point, taking the point of projection as the reference level.

Concept:

• In vacuum there is no air resistance, so the upward and downward journeys take equal times.
• Time to reach maximum height = total time ÷ 2.
• Potential energy: PE = mgh

Solution:

1. Time to reach maximum height: t = 1.2 ÷ 2 = 0.6 s

2. Initial velocity. At the top, v = 0, so: v = u − gt 0 = u − (10)(0.6) u = 6 m/s

3. Maximum height: h = ut − ½gt² h = (6)(0.6) − ½(10)(0.36) h = 3.6 − 1.8 = 1.8 m

Check: h = u²/2g = 36/20 = 1.8 m ✓

4. Convert the mass to kg: m = 20 g = 0.02 kg

5. Potential energy at maximum height: PE = mgh = 0.02 × 10 × 1.8 = 0.36 J

Answer: A (0.36 J)

Tip: Always convert grams to kilograms before using energy formulas. Options C and D (360 J and 720 J) come from forgetting that conversion and using 20 instead of 0.02. For "returns to the point of projection" questions, immediately halve the total time to get the time of ascent.