Thrown up at 30 m/s: height and total time

A ball is thrown straight up at 30 m/s (g = 10). Find the maximum height and the total time until it returns to the thrower's hand.

Step-by-step

1. h = v₀²/(2g) = 30²/(2×10) = 45
   The upward leg ends at speed 0, and v₀² = 2gh gives the maximum height
2. t = 2v₀/g = 2×30/10 = 6
   Rising and falling take equal times, so the total is 2 times the rise time

Why it works

At the highest point the speed is zero, so v₀² = 2gh gives the height; the rise and the fall take equal times, so the total time is twice the rise time.

Common mistakes

• Forgetting to double the rise time.
• Using g = 9.8 when the question says g = 10.

Related

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