Thrown up at 40 m/s: max height, total time

A ball is thrown straight up at 40 m/s (g = 10 m/s²). Find the maximum height it rises and the total time until it returns to the thrower's hand.

Step-by-step

1. h = v₀²/(2g) = 40²/(2×10) = 80
   The upward leg ends at speed 0, and v₀² = 2gh gives the maximum height
2. t = 2v₀/g = 2×40/10 = 8
   Rising and falling take equal times, so the total is 2 times the rise time

Why it works

Only gravity acts, so the speed drops by g every second and is zero at the highest point; the rise and the fall are mirror images, so the whole trip takes twice the rise time.

Common mistakes

• Using the full 40 m/s as the average speed over the whole rise.
• Reporting only the rise time and forgetting the fall back to the hand.

Related

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