Thrown up at 10 m/s: height and total time

A ball is thrown straight up at 10 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) = 10²/(2×10) = 5
   The upward leg ends at speed 0, and v₀² = 2gh gives the maximum height
2. t = 2v₀/g = 2×10/10 = 2
   Rising and falling take equal times, so the total is 2 times the rise time

Why it works

The speed drops by g every second, so the ball takes 1 s to stop; the height comes from v₀² = 2gh, and the symmetric fall doubles the time.

Common mistakes

• Saying the ball stops instantly because g equals the initial speed.
• Using h = v₀t with the full speed for the whole rise.

Related

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