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A ball is projected from ground into the air. At the height of $5 \mathrm{~m}$, its velocity is $\mathbf{v}=(5 \hat{i}+5 \hat{j}) \mathrm{ms}^{-1}$. The maximum height reached by the ball is (Acceleration due to gravity $=10 \mathrm{~m} \mathrm{~s}^{-2}$ )
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Verified Answer
The correct answer is:
$6.25 \mathrm{~m}$
The motion of projectile is shown below
To calculate height attained we consider only vertical motion.
Here, $u=5 \mathrm{~m} / \mathrm{s} ; a=-10 \mathrm{~m} / \mathrm{s}^2$
$v=0$
From $v^2-u^2=2 a s$
We have $0-5^2=2(-10) s$
$\Rightarrow \quad s=\frac{25}{20}=1.25 \mathrm{~m}$
Total maximum height reached by ball
$=5+1.25=6.25 \mathrm{~m}$.
To calculate height attained we consider only vertical motion.
Here, $u=5 \mathrm{~m} / \mathrm{s} ; a=-10 \mathrm{~m} / \mathrm{s}^2$
$v=0$
From $v^2-u^2=2 a s$
We have $0-5^2=2(-10) s$
$\Rightarrow \quad s=\frac{25}{20}=1.25 \mathrm{~m}$
Total maximum height reached by ball
$=5+1.25=6.25 \mathrm{~m}$.
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