Different Rules and Formulas Relating the Sides and Angles of the Triangle: Rules, Formulas

May 20, 202239 Insightful Publication

**Projectile Motion**: How does a ball move when it is thrown at some angle from the ground? At what angle should we throw the ball in order to cover the farthest distance? How to calculate the horizontal distance covered by the ball? What will be the maximum height of the projectile? What is the shape of the trajectory of a projectile? To find answers to these questions, let’s read further to get a better idea of what projectile motion is and how to analyse it.

Projectile motion is a two-dimensional motion in which the particle is thrown or projected at some angle with the horizontal. It can be analysed by breaking it into a combination of two one dimensional motions, that is, along the horizontal and along the vertical. It should be noted that acceleration only exists in the vertical direction if we chose the x-axis along the horizontal and the y-axis along the vertical.

When a body only moves in a single direction with respect to the chosen axis, then we call it motion in one dimension. Only the parameter of one particular direction changes while the rest remains the same. Example: A ball thrown vertically upwards is a motion in one dimension. The parameter along the horizontal does not change, while along the vertical, the velocity and the position of the ball change.

A bike moving on a straight line with some acceleration is also an example of motion in one dimension.

**Position:** It is the location of the particle with respect to the chosen reference point. It is denoted by \(\overrightarrow r \).

**Displacement:** It is the change in position of the particle with respect to the reference point. We calculate it by taking the difference between the final and the initial position. It is denoted by \(\overrightarrow S \).

**Distance: **It is the total length of the path taken by the particle.

**Velocity: **The rate of change of displacement with respect to time is known as velocity. It is denoted by \(\overrightarrow v \) or \(\overrightarrow u \)

**Acceleration: **The rate of change of velocity with respect to time is known as acceleration. It is denoted by \(\overrightarrow a \).

Negative acceleration is also sometimes referred to as retardation.

To ease the analysis of the motion of a particle in one dimension, three equations that describe the relationship between the parameters of the motion has been derived.

It is valid for a particle with uniform acceleration or zero acceleration.

The first equation of motion, \(\overrightarrow v = \,\overrightarrow u \, + \,\overrightarrow a t\)

Where,

\(\overrightarrow v \) is the initial velocity

\(\overrightarrow u \)the initial velocity

\(\overrightarrow a \) is the acceleration

\(t\) is the time taken.

The second equation of motion,

\(\overrightarrow s = \,\overrightarrow u t\, + \frac{1}{2}\,\overrightarrow a {t^2}\)

Where,

\(s\) is the displacement.

The third equation of motion,

\({v^2} – {u^2} = 2as\)

**The velocity of projection:** It is the velocity with which the body is projected. It is denoted by \(u\)

**The angle of projection:** It is the angle with the horizontal at which the body is projected. It is denoted by \(\theta\)

**Range:** It is the maximum distance covered by the projectile in the horizontal direction or x-axis.

The range of a projectile is given by,

\(R = \frac{{{u^2}\sin (2\theta )}}{g}\)

For maximum range, the angle of projection should be:

**Maximum height:** It is the maximum height reached by the projectile.

At the maximum height, the velocity in the vertical direction is zero.

The maximum height of the projectile is given by,

\(H = \frac{{{u^2}\sin (2\theta )}}{g}\)

**Time of flight:** It is the total time for which the ball remains in the air.

\(T = \frac{{2u\sin (\theta )}}{g}\)

Equation of path of the projectile is given by,

\(y = x\tan (\theta ) – \frac{{g{x^2}}}{{2{v^2}\cos (\theta )}}\)

From the above equation, we have the path of the projectile follows a parabolic trajectory.

Projectile motion up the plane and down the plane is a bit complex, but to make it easier, we rotate the axis so that the \(x\)-axis becomes along the inclined plane then the acceleration in both the horizontal and the vertical is non zero.

A basketball player should know the angle at which to throw the ball in order to put it in the basket to score.

Canon should know the angle at which it should fire to it its target.

*Q.1. A particle is projected on an inclined plane with speed, as shown in the figure. Find the range of the particle on the inclined plane.*

*Ans.*

Let the angle between the inclined plane and the horizontal is \(\alpha \).

Let the angle between the initial velocity and the horizontal is \(\theta \)

Let the x-axis be along the inclined plane.

Components of initial velocity,

\({u_x} = u\cos (\theta – \alpha )\)

\({u_x} = u\sin (\theta – \alpha )\)

And a component of acceleration,

\({a_x} = – g\sin \alpha \)

\({a_y} = – g\sin \alpha \)

Position along the inclined plane is given by,

\(x = u\cos (\theta – \alpha )t – \frac{1}{2}g\sin (\alpha ){t^2}\)

Position in the direction perpendicular to the inclined plane,

\(y = u\sin (\theta – \alpha )t – \frac{1}{2}g\cos (\alpha ){t^2}\)

For the time of flight, the position of the particle is zero along the direction perpendicular to the plane.

\(0 = u\sin (\theta – \alpha )T – \frac{1}{2}g\cos (\alpha ){T^2}\)

\(T = \frac{{2u\sin (\theta – \alpha )}}{{g\cos (\alpha )}}\)

Putting the value of time of flight in expression for position along the inclined plane.

\(R = u\cos (\theta – \alpha )\left( {\frac{{2u\sin (\theta – \alpha )}}{{g\cos (\alpha )}}} \right) – \frac{1}{2}g\sin (\alpha ){\left( {\frac{{2u\sin (\theta – \alpha )}}{{g\cos (\alpha )}}} \right)^2}\)

\(R = u\cos (45)\left( {\frac{{2u\sin (45)}}{{g\cos (30)}}} \right) – \frac{1}{2}g\sin (30){\left( {\frac{{2u\sin (45)}}{{g\cos (30)}}} \right)^2}\)

\(R = u\frac{1}{{\sqrt 2 }}\left( {\frac{{2u\frac{1}{{\sqrt 2 }}}}{{g\frac{{\sqrt 3 }}{2}}}} \right) – \frac{1}{2}g\frac{1}{2}{\left( {\frac{{2u\frac{1}{{\sqrt 2 }}}}{{g\frac{{\sqrt 3 }}{2}}}} \right)^2}\)

Solving we get,

\(R = \frac{{2{u^3}(\sqrt 3 – 1)}}{{3g}}\)

** Q.2. There are three paths and of a projectile projected from point as shown in figure**.

** Ans:** Let \(\alpha \) be the angle of projection with incline and \(\beta \) be the angle at which projectile strikes the plane.

be the angle at which projectile strikes the plane.

We know that

\(\tan \alpha = \frac{\rm{component}\;\rm{of}\;\rm{velocity}\;\rm{perpendicular}\;\rm{to}\;\rm{inclined}\;\rm{plane}}{\rm{component}\;\rm{of}\;\rm{velocity}\;\rm{parallel}\;\rm{to}\;\rm{the}\;\rm{inclined}\;\rm{plane}}\)

\(\tan (\alpha ) = \frac{{{v_3}}}{{{v_1}}}\)

\(\tan (\beta ) = \frac{{{v_4}}}{{{v_2}}}\)

Here is the magnitude of velocity component initially and finally along y-axis will be same, because \({S_y}\, = \,\,0\) and we know that for a projectile, the magnitude of the vertical component velocity is same if the position of the particle is same along the y-axis or the vertical.

Also, by using the third equation of motion,

\({v^2} – {u^2} = 2as\)

If the displacement is zero then the final and the initial velocities must be the same in magnitude

Hence,

\({v_3} = \,{v_4}\)

But the velocity component along x-axis goes on decreasing. So, \({v_2} = \,{v_1}\)

Hence, from above equations,

\(\tan (\beta ) > \tan (\alpha ) \Rightarrow \beta > \alpha \)

Hence, appropriate path should be \(b\).

Projectile motion is a two-dimensional motion in which the particle is thrown or projected at some angle with the horizontal.

Range: It is the maximum distance covered by the projectile in the horizontal direction or x-axis.

\(R = \frac{{{u^2}\sin (2\theta )}}{g}\)

Maximum height is the maximum height reached by the projectile.

\(H = \frac{{{u^2}\sin (\theta )}}{{2g}}\)

Time of flight: It is the total time for which the ball remains in the air.

\(T = \frac{{2u\sin (\theta )}}{g}\)

Equation of path of the projectile is given by,

\(y = x\tan (\theta ) – \frac{{g{x^2}}}{{2{v^2}\cos (\theta )}}\)

For maximum range is the angle of projection is \({45^ \circ }\).

*Q.1. What is a projectile motion?*** Ans:** Projectile motion is a two-dimensional motion in which the particle is thrown or projected at some angle with the horizontal.

*Q.2. What is the shape of the path of the projectile motion?*** Ans:** The equation of the path is given by,

\(y = x\tan (\theta ) – \frac{{g{x^2}}}{{2{v^2}\cos (\theta )}}\)

The above equation corresponds to a parabola; thus, the path of the projectile will be a parabola.

*Q.3. At what angle should be the projectile be projected to get maximum range?*** Ans:** If the angle of the inclined plane is \(\beta \)

The maximum range up the plane is given by,

\(\frac{{{u^2}}}{{g(1 + \sin \beta )}}\)

The maximum range down the plane will be given by,

\(\frac{{{u^2}}}{{g(1 – \sin \beta )}}\)

*Q.5. Will the equation of the projectile motion will change on the moon?*** Ans: **No, the expression of the equation of motion will remain the same only the value of the acceleration due to the gravity will change.

*We hope this detailed article on Projectile Motion helps you in your preparation. If you get stuck do let us know in the comments section below and we will get back to you at the earliest.*

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