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A ball is thrown at a speed of $20 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ with the horizontal. The maximum height reached by the ball is (Use $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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Verified Answer
The correct answer is:
$5 \mathrm{~m}$
Given,
Speed of ball $(u)=20 \mathrm{~m} / \mathrm{s}$
Angle $(\theta)=30^{\circ}$
We know that,
$$
\begin{aligned}
& H=\frac{u^2 \sin ^2 \theta}{2 g} \\
& H=\frac{(20)^2(\sin 30)^2}{2 \times 10} \\
& \Rightarrow \quad H=\frac{20 \times 20 \times\left(\frac{1}{2}\right)^2}{20} \\
& H=20 \times \frac{1}{4} \\
& \Rightarrow \quad H=5 \mathrm{~m} \\
&
\end{aligned}
$$
Speed of ball $(u)=20 \mathrm{~m} / \mathrm{s}$
Angle $(\theta)=30^{\circ}$
We know that,
$$
\begin{aligned}
& H=\frac{u^2 \sin ^2 \theta}{2 g} \\
& H=\frac{(20)^2(\sin 30)^2}{2 \times 10} \\
& \Rightarrow \quad H=\frac{20 \times 20 \times\left(\frac{1}{2}\right)^2}{20} \\
& H=20 \times \frac{1}{4} \\
& \Rightarrow \quad H=5 \mathrm{~m} \\
&
\end{aligned}
$$
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