MCQ
Figure shows a projectile thrown with speed $u=20 \,m / s$ at an angle $30^{\circ}$ with horizontal from the top of a building $40 \,m$ high. Then the horizontal range of projectile is ........... $m$
  • A
    $20 \sqrt{3}$
  • $40 \sqrt{3}$
  • C
    $40$
  • D
    $20$

Answer

Correct option: B.
$40 \sqrt{3}$
b
(b)

$S_y=u_y T+\frac{1}{2} g_y T^2$

$-40=4 \sin 30 T-\frac{1}{2} g T^2$

$-40=20 \times \frac{1}{2} T-5 T^2$

$-8=2 T-T^2$

$T^2-2 T-8=0$

$T^2-4 T+2 T-8=0$

$T=-2,4$

$R=u \cos \theta T=20 \times \frac{\sqrt{3}}{2} \times 4$

$R=40 \sqrt{3} \,m$

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