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Question: A projectile is projected with initial velocity \(\left( {6\hat i + 8\hat j} \right)m\,{s^{ - 1}}\) ...

A projectile is projected with initial velocity (6i^+8j^)m s−1\left( {6\hat i + 8\hat j} \right)m\,{s^{ - 1}} . If 10 ms−210\,m{s^{ - 2}} , then the horizontal angle is:
A. 9.6 m9.6\,m
B. 14.0 m14.0\,m
C. 4.8 m4.8\,m
D. 19.2 m19.2\,m

Explanation

Solution

Here, the initial velocity of the projectile is given in terms of x and y-coordinate. We will take these values of velocity in the horizontal and vertical direction in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta . Also, to calculate the horizontal angle, we will use the formula of the range of the projectile.

Complete step by step answer:
It is given in the question that a projectile is projected. A projectile is a substance that is thrown into the atmosphere by applying an external force. A projectile moves free under the influence of gravity. The gravity enables the projectile to move vertically in the atmosphere.
The initial velocity of the projectile, U=(6i^+8j^)m s−1U = \left( {6\hat i + 8\hat j} \right)m\,{s^{ - 1}}
Also, when the projectile is projected in the air, then the only force acting on the projectile is the acceleration due to gravity.
Therefore, acceleration due to gravity, g=10 ms−2g = 10\,m{s^{ - 2}}
Therefore, the x-component of the projectile is ux=6i^ ms−1{u_x} = 6\hat i\,m{s^{ - 1}}
Also, the y-component of the projectile is uy=8j^ ms−1{u_y} = 8\hat j\,m{s^{ - 1}}
Now, as we know that the x-component is the horizontal component, therefore, the velocity of the x-component is given by
ux=u cos⁡θ=6 ms−1{u_x} = u\,\cos \theta = 6\,m{s^{ - 1}}
Also, the y-component is the perpendicular component, therefore, the velocity of the projectile in the y-component is given by
uy=u sin⁡θ=8 ms−1{u_y} = u\,\sin \theta = 8\,m{s^{ - 1}}
Now, to calculate the horizontal angle we will use the range formula which is given by
R=u2 sin⁡2θgR = \dfrac{{{u^2}\,\sin 2\theta }}{g}
Now using the identity sin⁡2θ=2 sin⁡θ cos⁡θ\sin 2\theta = 2\,\sin \theta \,\cos \theta in the above equation, we get
R=u2(2sin⁡θ cos⁡θ)gR = \dfrac{{{u^2}\left( {2\sin \theta \,\cos \theta } \right)}}{g}
⇒ R=2(usin⁡θ)(ucos⁡θ)g\Rightarrow \,R = \dfrac{{2\left( {u\sin \theta } \right)\left( {u\cos \theta } \right)}}{g}
Now, putting the values in the above equation, we get
R=2(8)(6)10R = \dfrac{{2\left( 8 \right)\left( 6 \right)}}{{10}}
⇒ R=9610\Rightarrow \,R = \dfrac{{96}}{{10}}
∴ R=9.6 m\therefore \,R = 9.6\,m
Therefore, the horizontal angle is 9.6 m9.6\,m.

Hence, option A is the correct option.

Note: When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the center of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion. Air resistance to the motion of the body is to be assumed absent in projectile motion.