Thrown up at 80 m/s: height and time

A ball is thrown straight up at 80 m/s (g = 10 m/s²). Find the maximum height it can reach and the total time until it falls back to the thrower's hand.

Step-by-step

1. h = v₀²/(2g) = 80²/(2×10) = 320
   The upward leg ends at speed 0, and v₀² = 2gh gives the maximum height
2. t = 2v₀/g = 2×80/10 = 16
   Rising and falling take equal times, so the total is 2 times the rise time

Why it works

While rising the ball is in uniformly slowed motion with acceleration g downwards, and its speed reaches zero at the top; the upward and downward halves are symmetric, so the total time is double the rise time.

Common mistakes

• Thinking the ball still moves at 80 m/s at the highest point.
• Multiplying 80 m/s by the 8 s rise instead of using v₀² = 2gh.

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