Thrown up at 20 m/s: maximum height

A small ball is thrown vertically upward with an initial velocity of 20 m/s. Take g = 10 m/s². Find its maximum height and total time.

Step-by-step

1. h = v₀²/(2g) = 20²/(2×10) = 20
   The upward leg ends at speed 0, and v₀² = 2gh gives the maximum height
2. t = 2v₀/g = 2×20/10 = 4
   Rising and falling take equal times, so the total is 2 times the rise time

Why it works

Vertical upward motion can be divided into two symmetric stages: ascent and descent. The final velocity at the top is zero, and the initial velocity for the downward part is also zero. The time and height for both stages are equal. Therefore, to find the maximum height, only the velocity-displacement relationship is needed, and the total time is simply twice the ascent time. From an energy perspective, during the upward process, all kinetic energy converts into gravitational potential energy, yielding the same maximum height.

Common mistakes

• When calculating the total time, only the ascent time was computed, forgetting to multiply by two.
• Used the uniform acceleration displacement formula with initial velocity to calculate height, omitting the one-half factor.
• Mixed up the sign of gravitational acceleration with the direction of motion, resulting in a negative displacement that was directly taken as the height.

Related

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