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A ball is thrown vertically. It has a speed of $10 \mathrm{~m} / \mathrm{sec}$ when it has reached one half of its maximum height. How high does the ball rise? Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$.
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Verified Answer
The correct answer is:
$10 \mathrm{~m}$
Using equation of motion
$$
V^2=u^2-2 g h
$$
After reading max height velocity because zero.
$\begin{aligned} 0 & =(10)^2-2 \times 10 \times \frac{1}{2} \\ \Rightarrow \quad h & =\frac{200}{20}=10 \mathrm{~m}\end{aligned}$
$$
V^2=u^2-2 g h
$$
After reading max height velocity because zero.
$\begin{aligned} 0 & =(10)^2-2 \times 10 \times \frac{1}{2} \\ \Rightarrow \quad h & =\frac{200}{20}=10 \mathrm{~m}\end{aligned}$
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