The angle of projection to have maximum height and range in the ratio of 1:4 is
(A) $30{}^\circ$
(B) $45{}^\circ$
(C) $60{}^\circ$
(D) $75{}^\circ$
Answer
253.5k+ views
Hint We know that in a Projectile Motion, there are two simultaneous independent rectilinear motions: Along the x-axis: uniform velocity, responsible for the horizontal (forward) motion of the particle. Along y-axis: uniform acceleration, responsible for the vertical (downwards) motion of the particle. The projectile motion emphasizes one important aspect of constant acceleration that even constant acceleration, which is essentially unidirectional, is capable of producing two-dimensional motion. The basic reason is that force and initial velocity of the object are not along the same direction.
Complete step-by step answer
We know that a projectile is an object upon which the only force is gravity. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity. Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity.
In a projectile thrown at angle $\theta$ maximum height and range are given by:
Maximum Height ${{H}_{\max }}=\dfrac{{{U}^{2}}{{\sin }^{2}}\theta }{2g},\,Range\,R=\dfrac{{{U}^{2}}\sin 2\theta }{g}$
Now we have to find the ratio,
$\dfrac{{{H}_{\max }}}{R}=\dfrac{1}{4}=\dfrac{{{U}^{2}}{{\sin }^{2}}\theta \times g}{2g\times {{U}^{2}}\sin 2\theta }$
After the evaluation of the above expression we get the expression as:
$\dfrac{1}{4}=\dfrac{{{\sin }^{2}}\theta }{2\times 2\sin \theta \cos \theta }$
So, now we get that:
$\tan \theta =1$
$\Rightarrow \theta =45{}^\circ$
Hence, the correct answer is Option B.
Note We know that there are three main factors that affect the trajectory of an object or body in flight: the projection angle, magnitude of projection velocity and height of projection.
First let us know the meaning of projectile angle. The angle at which an observer draws an object to convert it from 3D to 2D plane such as real object image to a plane paper is known as angle of projection. The velocity with which the body is thrown is called the velocity of projection. The point from which the body is projected in the air is called a point of projection.
Complete step-by step answer
We know that a projectile is an object upon which the only force is gravity. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity. Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity.
In a projectile thrown at angle $\theta$ maximum height and range are given by:
Maximum Height ${{H}_{\max }}=\dfrac{{{U}^{2}}{{\sin }^{2}}\theta }{2g},\,Range\,R=\dfrac{{{U}^{2}}\sin 2\theta }{g}$
Now we have to find the ratio,
$\dfrac{{{H}_{\max }}}{R}=\dfrac{1}{4}=\dfrac{{{U}^{2}}{{\sin }^{2}}\theta \times g}{2g\times {{U}^{2}}\sin 2\theta }$
After the evaluation of the above expression we get the expression as:
$\dfrac{1}{4}=\dfrac{{{\sin }^{2}}\theta }{2\times 2\sin \theta \cos \theta }$
So, now we get that:
$\tan \theta =1$
$\Rightarrow \theta =45{}^\circ$
Hence, the correct answer is Option B.
Note We know that there are three main factors that affect the trajectory of an object or body in flight: the projection angle, magnitude of projection velocity and height of projection.
First let us know the meaning of projectile angle. The angle at which an observer draws an object to convert it from 3D to 2D plane such as real object image to a plane paper is known as angle of projection. The velocity with which the body is thrown is called the velocity of projection. The point from which the body is projected in the air is called a point of projection.
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