One second after projection, a projectile is travelling in a direction inclined at $45^{\circ}$ to…

One second after projection, a projectile is travelling in a direction inclined at $45^{\circ}$ to horizontal. After two more seconds it is travelling horizontally. Then the magnitude of velocity of the projectile is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
  1. $10 \sqrt{13} \mathrm{~ms}^{-1}$
  2. $11 \mathrm{~ms}^{-1}$
  3. $10 \sqrt{2} \mathrm{~ms}^{-1}$
  4. $20 \mathrm{~ms}^{-1}$

Solution

When projectile travels horizontally, then $t=\frac{T}{2}$ $\therefore \frac{\mathrm{T}}{2}=3 \Rightarrow \frac{\mathrm{u} \sin \theta}{\mathrm{~g}}=3$ $\mathrm{u}, \mathrm{u} \sin \theta=30$ After time, $\mathrm{t}=1 \mathrm{~s}, \tan 45^{\circ}=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}} \Rightarrow \mathrm{v}_{\mathrm{y}}=\mathrm{v}_{\mathrm{x}}$ $\begin{aligned} & \Rightarrow u \sin \theta-g t=u \cos \theta \\ & \Rightarrow 30-10 \times 1=u \cos \theta\end{aligned}$ $\begin{aligned} & \therefore u_x=u \cos \theta=20 \\ & \therefore u=\sqrt{u_x^2+u_y^2}=\sqrt{(30)^2+(20)^2}=10 \sqrt{13} \mathrm{~m} / \mathrm{s}\end{aligned}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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