Projectile Motion

Projectile Motion is the motion of a body under the influence of gravity only. A body which moves with constant velocity in horizontal direction and experience uniform acceleration due to gravity (9.8 m/s²) in vertically downward direction is called projectile. So, this type of motion is considered as projectile motion. Projectile motion is a 2D motion.

For e.g. dropping a bomb from an aeroplane, firing a bullet from a gun etc. The path followed by a projectile is called trajectory. The projectile always follows a parabolic path(trajectory).

Proof that trajectory is parabola:

For\quad motion\quad in\quad Y- axis;\\y=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }\\\quad =u\sin { \theta .t-\frac { 1 }{ 2 }  } g{ t }^{ 2 }............(i)\\ For\quad motion\quad in\quad X-axis;\\x=ut\\ \quad =u\cos { \theta .t } \\\Rightarrow t=\frac { x }{ u\cos { \theta  }  }..................(ii)\\ Replacing \quad t \quad in\quad equation (i)\\y=u\sin { \theta  } .\frac { x }{ u\cos { \theta  }  } -\frac { 1 }{ 2 } g.\frac { { x }^{ 2 } }{ { u }^{ 2 }\cos ^{ 2 }{ \theta  }  } \\ \quad =\tan { \theta .x-\frac { g }{ 2{ u }^{ 2 }\cos ^{ 2 }{ \theta  }  }  } .x^{ 2 }\\ \quad =ax-b{ x }^{ 2 }\quad(eq.\quad of\quad parabola)\\

Equations for projectile motion.

While discussing the projectile motion considering other forces like friction, drag etc. complicates the problem.so, we will neglect the drag force and other additional forces.

i.Time Of Flight:

If you through a ball vertically upward with an initial velocity ‘u’ the time require to reach the maximum height is \frac { u }{ g } sec. Because the velocity is ‘u’ and the velocity decreases per second is g\frac { m }{ s } .

Again to return to the ground it requires \frac { u }{ g }sec.

So,total time taken is \frac { u }{ g } +\frac { u }{ g } =\frac { 2u }{ g } sec.

But the vertical component is ;u\sin { \theta } \\

ii.Maximum height:

We have the equation of motion as ;{ v }^{ 2 }-{ u }^{ 2 }=2as. But for maximum height , v=0

So, s=\frac { { -u }^{ 2 } }{ 2g }.If we take the magnitude only then  s=\frac { { u }^{ 2 } }{ 2g }.

But the vertical component is; u\sin { \theta }

iii. Maximum Range:

R=uT;\\But;u=u\cos { \theta }\\\quad \quad T=\frac { 2u\sin { \theta } }{ g }\\So,R=u\cos { \theta \ast } \frac { 2u\sin { \theta } }{ g }

FAQS:-

1.Dos the formula “R=uT” work for the range of the projectile?

Of course, consider the following animation where a rider jumps from his skateboard and lands on the same after some time. Here the velocity of the skateboard is equal to the horizontal velocity of the rider i.e u\cos { \theta }. Instead of considering the parabolic path of the body we consider the straight line path of the skateboard and apply R=uT. During the whole journey rider always remains on the head of the skateboard.

2. Why there is no acceleration in horizontal direction?

With whatever speed you throw an object into the air once it is released from your hand now it moves under the influence of gravity only. It is a case of free fall. The gravity acts in vertically downward direction only and there is no opposing force in the horizontal direction. Also remember that we neglect the air resistance.