A ball is thrown vertically upward with velocity 30 m/s. Taking g = 10 m/s², calculate the maximum height and total flight time.

## **The reasoning** When the ball reaches maximum height, its velocity becomes **zero**. Using the principle of **conservation of energy** (or the kinematic equation): **For maximum height:** v² = u² − 2gh At the top, v = 0, u = 30 m/s, g = 10 m/s² 0 = (30)² − 2(10)h 0 = 900 − 20h h = **45 m** **For total flight time:** What goes up must come down! Time to go up = time to come down. Using v = u − gt: 0 = 30 − 10t t = 3 s (time to reach the top) Total flight time = 3 + 3 = **6 s** --- ## **Why the wrong options tempt you** **Option B (45 m, 3 s)** — You calculated height correctly but forgot the ball must come back down! 3 s is only one-way time. **Option C (90 m, 3 s)** — You used h = ut instead of the correct formula, and made the same time mistake. **Option D (90 m, 6 s)** — Right time logic, but wrong height calculation. --- ## **Quick takeaway** For vertical projection: **total time = 2 × time to maximum height**, and always use v² = u² − 2gh for height when final velocity is zero!