A body is projected vertically upwards with a speed of 10 ms^ -1 from a point 2m above the ground. Calculate…

Analyze the question: The body is thrown upward at 10 m/s from 2 m above the ground. It rises, stops, falls back past its starting point, and then falls the extra 2 m to the ground. We need the total time from launch to ground impact.

Concept: Use the equation of motion under gravity, taking upward as positive:

s = ut + ½at²

The final position is 2 m below the starting point, so the displacement is s = −2 m, with a = −10 m/s² and u = +10 m/s.

Solution:

−2 = 10t − 5t²

Rearranging:

5t² − 10t − 2 = 0

Using the quadratic formula:

t = [10 ± √(100 + 40)] / 10 = [10 ± √140] / 10

√140 ≈ 11.83, so:

t = (10 + 11.83)/10 ≈ 2.18 s

The negative root (t ≈ −0.18 s) is rejected because time cannot be negative.

Check by parts:

• Time to reach the top: t = u/g = 1 s
• Max height above the start: u²/2g = 5 m, so the maximum height is 7 m above the ground
• Fall time from 7 m: t = √(2×7/10) = √1.4 ≈ 1.18 s
• Total: 1 + 1.18 = 2.18 s ✓

Answer: C. 2.18 s

Exam tip: When a body starts above the ground, don't just double the time to the top (that would give 2 s, the trap option B). That only gets you back to the starting height. You must add the extra time to fall the remaining distance, or use s = ut + ½at² directly with the correct sign for displacement.