When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with horizontal, it can travel a distance x1 along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x2 distance. Then x1: x2 will be:

Question

When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with horizontal, it can travel a distance x1 along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x2 distance. Then x1: x2 will be: (A) 1:√3 (B) 1: 2√3 (C) 1:√2 (D) √2:1

Tardigrade Admin · Accepted Answer

( Stopping distance ) x 1=( u 2/2 g sin 60°) (Stopping distance) x 2=( u 2/2 g sin 30°) ⇒ ( x 1/ x 2)=( sin 30°/ sin 60°)=(1 × 2/2 × √3)=1: √3

$1 : 3$

$1 : 23$

$1 : 2$

$2 : 1$

$($ Stopping distance $) x_{1} = \frac{u ^{2}}{2 g s i n 6 0 ^{\circ}}$
(Stopping distance) $x_{2} = \frac{u ^{2}}{2 g s i n 3 0 ^{\circ}}$
$\Rightarrow \frac{x _{1}}{x _{2}} = \frac{s i n 3 0 ^{\circ}}{s i n 6 0 ^{\circ}} = \frac{1 \times 2}{2 \times 3} = 1 : 3$

1:3​

1:23​

1:2​

2​:1

( Stopping distance )x1​=2gsin60∘u2​ (Stopping distance) x2​=2gsin30∘u2​ ⇒x2​x1​​=sin60∘sin30∘​=2×3​1×2​=1:3​

$1 : 3$

$1 : 23$

$1 : 2$

$2 : 1$

$($ Stopping distance $) x_{1} = \frac{u ^{2}}{2 g s i n 6 0 ^{\circ}}$
(Stopping distance) $x_{2} = \frac{u ^{2}}{2 g s i n 3 0 ^{\circ}}$
$\Rightarrow \frac{x _{1}}{x _{2}} = \frac{s i n 3 0 ^{\circ}}{s i n 6 0 ^{\circ}} = \frac{1 \times 2}{2 \times 3} = 1 : 3$