Q124 / 160 · AP EAMCET 2023 (17 May Shift 1) · Physics

At any instant ' $\mathrm{t}$ ' the vertical distance ' $\mathrm{Y}$ ' and horizontal distance ' $\mathrm{X}$ ' of a projectile are given by $2 \mathrm{Y}=6 \mathrm{t}-\mathrm{gt}^2$ and $X=4 \mathrm{t}$. The initial velocity of projectile is ( $\mathrm{X}$ and $\mathrm{Y}$ are in metre and $\mathrm{t}$ is in second)

  • A
    $3 \mathrm{~ms}^{-1}$
  • B
    $4 \mathrm{~ms}^{-1}$
  • Correct answer
    C
    $5 \mathrm{~ms}^{-1}$
  • D
    $6 \mathrm{~ms}^{-1}$

Solution

$\begin{aligned} & x=4 t \\ & V_x=\frac{d x}{d t}=4 \\ & Y=3 t-\frac{g t^2}{2} \\ & V_y=\frac{d y}{d t}=3-g t \\ & \text { at } t=0 V_y=3\end{aligned}$ $\begin{aligned} & V_i=\sqrt{V_x^2+V_y^2} \\ & =\sqrt{4^2+3^2}=5 \mathrm{~m} / \mathrm{s}\end{aligned}$

Hence, the correct answer is option: 'C'.

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