Question:
The range of a particle projected at an angle of 15? with the horizontal is 1.5 km. Its range when projected with the same velocity at an angle of 45? with the horizontal is
The range of a particle projected at an angle of 15? with the horizontal is 1.5 km. Its range when projected with the same velocity at an angle of 45? with the horizontal is
Updated On: May 12, 2024
- 1.5 km
- 3 km
- 0.75 km
- 4.5 km
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The Correct Option is B
Solution and Explanation
We know, $R = \frac{u^2 \, \sin \, 2\theta}{g}$
Here, $ \theta = 15^\circ, R = 1.5 \, km = 1500 \, m $
$ \therefore \:\:\: 1500 = \frac{u^2 }{g} \sin 30^\circ$ or , $ \frac{u^2}{g} = 3000 \, m$
Now for same velocity and $\theta = 45?$, range of the projectile would be,
$R' = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g} = 3000 \, m = 3 \, km $
Here, $ \theta = 15^\circ, R = 1.5 \, km = 1500 \, m $
$ \therefore \:\:\: 1500 = \frac{u^2 }{g} \sin 30^\circ$ or , $ \frac{u^2}{g} = 3000 \, m$
Now for same velocity and $\theta = 45?$, range of the projectile would be,
$R' = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g} = 3000 \, m = 3 \, km $
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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