Q132 / 170 · Manipal (MET) 2016 · Physics

A projectile has the maximum range 500 m . If the projectile is thrown up an inclined plane of $30^{\circ}$ with the same (magnitude) velocity, the distance covered by it along the inclined plane will be

  • A
    250 m
  • Correct answer
    B
    500 m
  • C
    400 m
  • D
    1000 m

Solution

Since, range is maximum, therefore $\theta=45^{\circ}$. Hence, using $R=u^2 \sin 2 \theta / g$. We find $500=\frac{u^2}{g}$, i.e. $u=[500 g]^{\frac{1}{2}}$. Distance covered along the inclined plane can be obtained using relation $v^2-v_0^2=2 a x$ $\Rightarrow \quad 0-u^2=2 \times\left(-g \sin 30^{\circ}\right) \times x$ $\left[\because a=-g \sin 30^{\circ}\right]$ $\therefore \quad x=\frac{u^2}{g}=500 \mathrm{~m}$

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

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