Maximum Height in Vertical Throw – Rankers Physics
Topic: Kinematics
Subtopic: Motion Under Gravity

Maximum Height in Vertical Throw

A ball is thrown vertically upward. It has a speed of \(10\text{ m/s}\) when it has reached one half of its maximum height. How high does the ball rise? (Taking \('g' = 10\text{ m/s}^2\)\)

(2005)

\(6\text{ m}\)
\(10\text{ m}\)
\(14\text{ m}\)
\(18/text{ m}\)

Solution:

Concept: Vertical motion under gravity. Let \(H\) be maximum height, \(u\) be initial velocity.
Formula: \(v^2 = u^2 - 2gh\). At \(H\), \(v=0 \Rightarrow u^2 = 2gH\).
At \(H/2\), \(10^2 = u^2 - 2g(H/2) = u^2 - gH\).
Substitute \(u^2 = 2gH\): \(100 = 2gH - gH = gH\).
Given \(g=10\text{ m/s}^2\), so \(100 = 10H \Rightarrow H = 10\text{ m}\).

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