Q76 / 100 · BITSAT 2023 (Memory Based Paper 2) · Physics
A projectile is projected at $30^{\circ}$ from horizontal with initial velocity $40 \mathrm{~ms}^{-1}$. The velocity of the projectile at $\mathrm{t}=2 \mathrm{~s}$ from the start will be: (Given $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
- Correct answerA$20 \sqrt{3} \mathrm{~ms}^{-1}$
- B$40 \sqrt{3} \mathrm{~ms}^{-1}$
- C$20 \mathrm{~ms}^{-1}$
- D
Zero
Solution
Given,
Initial velocity of projectile, $u=40 \mathrm{~m} / \mathrm{s}$
Angle, $\theta=30^{\circ}$
Time of flight
$\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=\frac{2 \times 40 \times 1}{10 \times 2}=45\left(\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$
It means projectile is at maximum height at $t=$ $2 \mathrm{~s}$. At maximum height vertical component of velocity is zero.
$\begin{aligned}<br />& \text { Velocity at } \mathrm{t}=2 \mathrm{~s}=\mathrm{V}_{\mathrm{x}}=\mathrm{u} \cos \theta=40 \cos 30^{\circ} \\<br />& =20 \sqrt{3} \mathrm{~ms}^{-1} \text {. } \\<br />&<br />\end{aligned}$
Initial velocity of projectile, $u=40 \mathrm{~m} / \mathrm{s}$
Angle, $\theta=30^{\circ}$
Time of flight
$\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=\frac{2 \times 40 \times 1}{10 \times 2}=45\left(\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$
It means projectile is at maximum height at $t=$ $2 \mathrm{~s}$. At maximum height vertical component of velocity is zero.
$\begin{aligned}<br />& \text { Velocity at } \mathrm{t}=2 \mathrm{~s}=\mathrm{V}_{\mathrm{x}}=\mathrm{u} \cos \theta=40 \cos 30^{\circ} \\<br />& =20 \sqrt{3} \mathrm{~ms}^{-1} \text {. } \\<br />&<br />\end{aligned}$
Hence, the correct answer is option: 'A'.