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A small ball is thrown at an angle $45^{\circ}$ to the horizontal with an initial velocity of $2 \sqrt{2} \mathrm{~m} / \mathrm{s}$. The magnitude of mean velocity averaged over the first $2 \mathrm{~s}$ is [talse, acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
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Verified Answer
The correct answer is:
$8.2 \mathrm{~m} / \mathrm{s}$
Horizontal displacement in 2 seconds,
$$
x=u_x \times t=2 \sqrt{2} \cos 45^{\circ} \times 2=4 \mathrm{~m}
$$
Vertical displacement in 2 seconds,
$$
\begin{aligned}
y & =u_{y t}+\frac{1}{2} a_{y t} 2 \\
& =2 \sqrt{2} \sin 45^{\circ} \times 2-\frac{1}{2} \times 10 \times 4=4-20=-16 \mathrm{~m}
\end{aligned}
$$
Total displacement of particle in first 2 seconds
$$
=\sqrt{x^2+y^2}=\sqrt{4^2+16^2}=16.49 \mathrm{~m}
$$
$$
\begin{aligned}
\text { Average Velocity } & =\frac{\text { Total displacement }}{\text { Total time }} \\
& =\frac{16.49}{2}=8.24 \mathrm{~ms}^{-1} \approx 8.2 \mathrm{~ms}^{-1}
\end{aligned}
$$
$$
x=u_x \times t=2 \sqrt{2} \cos 45^{\circ} \times 2=4 \mathrm{~m}
$$
Vertical displacement in 2 seconds,
$$
\begin{aligned}
y & =u_{y t}+\frac{1}{2} a_{y t} 2 \\
& =2 \sqrt{2} \sin 45^{\circ} \times 2-\frac{1}{2} \times 10 \times 4=4-20=-16 \mathrm{~m}
\end{aligned}
$$
Total displacement of particle in first 2 seconds
$$
=\sqrt{x^2+y^2}=\sqrt{4^2+16^2}=16.49 \mathrm{~m}
$$
$$
\begin{aligned}
\text { Average Velocity } & =\frac{\text { Total displacement }}{\text { Total time }} \\
& =\frac{16.49}{2}=8.24 \mathrm{~ms}^{-1} \approx 8.2 \mathrm{~ms}^{-1}
\end{aligned}
$$
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