Stone thrown up at 15 m/s: height and time

A stone is thrown straight up at 15 m/s (g = 10). Find the maximum height and the total time until it comes back to the thrower's hand.

Step-by-step

1. h = v₀²/(2g) = 15²/(2×10) = 11.25
   The upward leg ends at speed 0, and v₀² = 2gh gives the maximum height
2. t = 2v₀/g = 2×15/10 = 3
   Rising and falling take equal times, so the total is 2 times the rise time

Why it works

The stone rises until its speed reaches zero, and the height follows from v₀² = 2gh; because the motion is symmetric, the total time is twice the rise time.

Common mistakes

• Rounding the rise time to 2 s and then doubling it.
• Thinking the return speed must be smaller than 15 m/s.

Related

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