Question:
A plane is inclined at an angle $\alpha = 30^{\circ}$ with a respect to the horizontal. A particle is projected with a speed $u = 2 ms^{-1}$ from the base of the plane, making an angle $q = 15^{\circ}$ with respect to the plane as shown in the figure. The distance from the base, at which the particle hits the plane is close to :
(Take $g = 10 \; ms^{-2}$)
A plane is inclined at an angle $\alpha = 30^{\circ}$ with a respect to the horizontal. A particle is projected with a speed $u = 2 ms^{-1}$ from the base of the plane, making an angle $q = 15^{\circ}$ with respect to the plane as shown in the figure. The distance from the base, at which the particle hits the plane is close to :
(Take $g = 10 \; ms^{-2}$)
Updated On: Apr 28, 2025
- 14 cm
- 20 cm
- 18 cm
- 26 cm
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The Correct Option is B
Solution and Explanation
$t = \frac{2\times2\times\sin 15^{\circ}}{g \cos 30^{\circ}} $
$ S= 2 \cos 15^{\circ} \times t - \frac{1}{2} g \sin30^{\circ} t^{2} $
Put values and solve
S $\simeq$ 20cm
$ S= 2 \cos 15^{\circ} \times t - \frac{1}{2} g \sin30^{\circ} t^{2} $
Put values and solve
S $\simeq$ 20cm
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration