# Projectile

A projectile is any object having a given initial velocity and moving along a path determined by the force of gravity and frictional resistance of the air. In an idealized situation, the object is assumed so small that it may be regarded as a particle. The air resistance is neglected. And the particle is supposed to be closed to the surface of the earth so that the acceleration due to gravity is almost constant.

## Projectile Fired By Making An Angle With Horizontal

​Consider a projectile is fired with velocity u from a point O (origin) by making an angle θ with he horizontal. OX is X-axis and OY is Y-axis. Then, the velocity u can be resolved into two components; ucosθ along X-axis and usinθ along Y-axis.

Here,

• ucosθ remains constant throughout the motion of the projectile.
• usinθ continuosly varies due to acceleration due to gravity.

And, the actual path followed by the projectile is known as trajectory.

### The path followed by the projectile is parabolic.

​The motion of the projectile along Y-axis is, $y=u_y t -\frac{1}{2}gt^2$ $y=usinθ\cdot t -\frac{1}{2}gt^2$ and, the motion of the projectule along X-axis is, $x=ucosθ ×t$ $∴t=\frac{x}{ucosθ}$ Thus, we have, $y=usinθ\cdot \frac{x}{ucosθ}-\frac{1}{2}g\left(\frac{x}{ucosθ}\right)^2$ $y=tanθ\cdot x-\left(\frac{g}{2u²cos²θ}\right)x^2$ This equation can be written as, $y=ax-bx^2$ This is a parabolic equation. Thus, the path followed by the projectile is parabolic.

Let the mass of the projected particle be m and suppose it is at the point P(x, y) at the end of time t.

Let Fx be the component of the force of gravity on the particle along X-axis and ax be the corresponding component of the acceleration due to gravity, then, by Newton’s second law of motion, $a_x=\frac{F_x}{m}=0$ Similarly, if Fy is the component of the force of gravity on the particle along Y-axis and ay be the corresponding component of the acceleration due to gravity, then, $a_y=\frac{F_y}{m}=\frac{-mg}{m}=-g$ Let v be the velocity of the particle at P(x, y) and α be its inclination to the horizontal. Then, $a_x=\frac{vcosα-ucosθ}{t}=0$ $∴vcosα=ucosθ$ and, $a_y=\frac{vsinα-usinθ}{t}=-g$ $∴vsinα=usinθ-gt$

### Velocity at any height

We have, horizontal component of velocity v at point P, $vcosα=ucosθ$ and, vertical component of velocity v at point P, $vsinα=usinθ-gt$ Squaring and adding these two components, we get, $v^2(cos^2α+sin^2α)=u^2(cos^2θ+sin^2θ)-2usinθgt+g^2t^2$ $v^2=u^2-2usinθgt+g^2t^2$ $v^2=u^2-2g(usinθt-\frac{1}{2}gt^2)$ $∴v^2=u^2-2gy$ Where, y is the vertical height of the particle above OX. Also, $\frac{vsinα}{vcosα}=\frac{usinθ-gt}{ucosθ}$ $tanα=\frac{usinθ-gt}{ucosθ}$ $∴α=tan^{-1}\left(\frac{usinθ-gt}{ucosθ}\right)$ Thus, from these equations, we can find both the magnitude and direction of velocity of the projectile at a given point or time.

### Time of Flight

Time of flight is the total time taken by the projectile from its point of projection to its point of hitting the ground.
​We have, $y=u_yt-\frac{1}{2}gt^2$ $y=usinθ\cdot t-\frac{1}{2}gt^2$ When t=T(Time of flight), then y = 0, $0=usinθ\cdot t-\frac{1}{2}gt^2$ $\frac{1}{2}gT^2=usinθT$ $T=\frac{2usinθ}{g}$ This is the time of flight of the projectile.

### Maximum Height Attained

​The projectile, during its motion, attains a maximum height.
​We have, $v_y^2=u_y^2-2gh$ $\text{When }h=h_{max}\text{ (maximum height), then }v_y=0$$0=u_y^2-2gh_{max}$ $2gh_{max}=u^2sin²θ$ $h_{max}=\frac{u^2sin²θ}{2g}$ This gives the maximum height attained by the projectile.

### Horizontal Range

​The total distance covered by the projectile from its point of projection to its point of hitting the ground is known as horizontal range.
​The distance covered by the projectile along X-axis is, $x=u_xt$ $x=ucosθ\cdot t$ $\text{When }t=T\text{ (Time of flight), then } x=R \text{ (horizontal range)}$ $∴R=ucosθ\cdot T$ We have, $T=\frac{2usinθ}{g}$ $∴R=ucosθ\cdot \frac{2usinθ}{g}$ $R=\frac{2u²sinθcosθ}{g}$ $R=\frac{u²sin2θ}{g}$ This gives the horizontal range of the projectile. When θ=45°, then the horizontal range is maximum. i.e. $R=\frac{u²sin(2×45)}{g}$ $R=\frac{u²sin90}{g}$ $R=\frac{u²}{g}$

## Projectile Fired Horizontally From The Top Of A Tower

​Let us consider a projectile fired from the top of a tower of hight h with initial velocity u. Then, the component of initial velocity along X-axis. $u_x = u$ and, the component of initial velocity along Y-axis. $u_y = 0$ Then, the X-componet of initial velocity remains constant throughout the motion. While, the Y-component of initial velocity changes with respect to time due to acceleration due to gravity.

### Path of the Projectile

Suppose at any time t, the projectile is at a point P(x,y). Then, the distance x covered by the projectile in time t is, $x=u_xt$ $x=ut\text{ ____(1)}$ and, the distance y covered by the projectile in time t is, $y=u_yt+\frac{1}{2}gt^2$ $y=0+\frac{1}{2}gt^2$ $y=\frac{1}{2}gt^2$ From (1), we have, $t=\frac{x}{u}$ $∴y=\frac{1}{2}g\frac{x^2}{u^2}$ $y=\left(\frac{g}{2u^2}\right)x^2$ $y=bx^2$ This equation is parabolic. So, the path followed by the projectile is parabola.

### Velocity of Projectile at P

The horizontal component of the velocity at time t is given by, $v_x=u_x=u$ The vertical component of the velocity at time t is given by, $v_y=u_y+gt$ $v_y=gt$ $∴\text{Resultant Velocity (v)}=\sqrt{v_x^2+v_y^2}$ =\sqrt{u^2+g^2t^2\] and, the angle of resultant velocity, $tanθ=\frac{v_y}{v_x}$ $tanθ=\frac{gt}{u}$ $θ=tan⁻¹\left(\frac{gt}{u}\right)$

### Time of Flight

​The time taken by the projectile to reach to the ground is known as time of flight. We have, $y=\frac{1}{2}gt^2$ When y=h (height of tower), then t=T(time of flight) $∴h=\frac{1}{2}gT^2$ $∴T=\sqrt{\frac{2h}{g}}$ ​Velocity of the projectile when it reaches the ground, We have, $v=\sqrt{u^2+g^2t^2}$ When the projectile reaches the ground, t=T, $v=\sqrt{u^2+g^2T^2}$ $v=\sqrt{u^2+g^2\frac{2h}{g}}$ $v=\sqrt{u^2+2gh}$ Also, direction of velocity at the ground, $tanθ=\frac{gT}{u}$ $tanθ=\frac{g\sqrt{\frac{2h}{g}}}{u}$ $tanθ=\frac{\sqrt{2gh}}{u}$ $θ=tan⁻¹\left(\frac{\sqrt{2gh}}{u}\right)$

### Horizontal Range

Horizontal Range (R) is the horizontal distance covered by the projectile in the time of flight. We have, $x=ut$ When t=T, x=R, $R=uT$ $R=u\sqrt{\frac{2h}{g}}$

Q. For a given velocity of projection and a given horizontal range, prove that there are, in general, two directions of projection which are equally inclined to the direction of maximum range.

Let u be the velocity of projection, R be the horizontal range and θ be the angle of projection. Then, $R=\frac{u^2}{g}sin2θ$ $R=\frac{u^2}{g}sin(π-2θ)$ $∴R=\frac{u^2}{g}sin2\left(\frac{π}{2}-θ\right)$ i.e. the value of the range remains same if θ is replaced by π/2-θ.

Moreover, these two directions are equally inclined to the horizontal and vertical axes respectively. Hence, the direction of maximum range (45°) bisects the angle between these two directions.

Q. A ball is thrown from the top of a building towards a tall building 50√3 m away. The initial velocity of the ball is 20 m/s at 30° above the horizontal. How far above or below its original level will the ball strike the opposite wall? (Take g = 10 m/s)

Let the ball thrown away from the top A of a building strike another building at a point B after t seconds. Let y be the distance of B from the original level. Then, horizontal component of initial velocity $=20cos30=10\sqrt{3} \text{ ms}^{-1}$ and, vertical component of initial velocity $=20sin30=10 \text{ ms}^{-1}$ and, for the horizontal motion, $50\sqrt{3}=10\sqrt{3}×t$ $t=5\text{ secs}$ During these 5 seconds, the vertical distance covered, $y=10.5+\frac{1}{2}(-10)×25=-75 \text{ m}$ The negative sign shows that it strikes the opposite wall 75 m below the original level.

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