Q124 / 160 · AP EAMCET 2022 (05 Jul Shift 1) · Physics

A projectile is launched from the ground, such that it hits a target on the ground which is $90 \mathrm{~m}$ away. The minimum velocity of projectile to hit the target is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

  • A
    10. $\mathrm{ms}^{-1}$
  • B
    $16 \mathrm{~ms}^{-1}$
  • C
    $60 \mathrm{~ms}^{-1}$
  • Correct answer
    D
    $30 \mathrm{~ms}^{-1}$

Solution

Since, launched projectile hit the target on the ground $90 \mathrm{~m}$ away hence, range, $R=90 \mathrm{~m}$. For minimum velocity $\left(u_{\min }\right)$ of projectile, range should be maximum. i.e. $\theta=45^{\circ}$ $R=\frac{u^2 \sin 2 \theta}{g}$ $\begin{array}{ll}\Rightarrow & 90=\frac{u_{\min }^2 \sin 2 \times 45^{\circ}}{10} \Rightarrow 900=u_{\min }^2 \\ \Rightarrow & u_{\min }=\sqrt{900}=30 \mathrm{~m} / \mathrm{s}\end{array}$

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

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