Question 174: Motion in 2D & Vectors

Question

When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60
o
with horizontal, it can travel a distance x
1
along the plane. But when the inclination is decreased to 30
o
and the same object is shot with the same velocity, it can travel x
2
distance. Then x
1
: x
2
will be :

Accepted Answer

Assume initial velocity = u
In first case, Moved distance, (x
1
) = $${{{u^2}} \over {2g\sin {{60}^o}}}$$
In second case, Moved distance, (x
2
) = $${{{u^2}} \over {2g\sin {{30}^o}}}$$
$$ 	herefore $$ $${{{x_1}} \over {{x_2}}} = {{\sin {{30}^o}} \over {\sin {{60}^o}}}$$ = $${{1 	imes 2} \over {2 	imes \sqrt 3 }}$$ = $${1 \over {\sqrt 3 }}$$

Answer

1 : $$\sqrt 2 $$

Answer

$$\sqrt 2 $$ : 1

Answer

1 : $$\sqrt 3 $$

Answer

1 : 2$$\sqrt 3 $$

Motion in 2D & Vectors

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