Thrown up at 50 m/s: height and total time

A ball is thrown straight up at 50 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) = 50²/(2×10) = 125
   The upward leg ends at speed 0, and v₀² = 2gh gives the maximum height
2. t = 2v₀/g = 2×50/10 = 10
   Rising and falling take equal times, so the total is 2 times the rise time

Why it works

The ball slows at g on the way up, so the height comes from v₀² = 2gh; the upward and downward halves are mirror images, so the total time is twice the rise time.

Common mistakes

• Reporting only the rise time and forgetting the fall.
• Dividing the height by g instead of using v₀² = 2gh.

Related

vertical-throw pendulum-period vertical-throw-40ms-max-height