# Projectile Motion Formula

One often experiences many kinds of motions in our daily life. Projectile motion is one among them. A projectile is some body thrown in space or air. The Curved path with which the projectile travels is what is termed as trajectory. The free fall motion of any object in a horizontal path with constant velocity is Projectile Motion.

Projectile Motion Formula (trajectory formula) is articulated as

$Horizontal\;&space;distance,x=V_{x}t$

$Horizontal\;&space;distance,V_{x}=V_{x_{o}}$

$Vertical\;&space;distance,y=V_{yo}t-\frac{1}{2}gt^{2}$

$Vertical\;&space;Velocity,V_{y}=V_{yo}-gt$

Where,
the velocity along the x-axis is V,
the initial velocity along the x-axis is Vxo ,
the velocity along the y-axis is Vy ,
the initial velocity along the y-axis is Vyo .
acceleration due to gravity is g, and
the time taken is t

Equations associated to the trajectory motion (projectile motion) are articulated as,

$Time&space;of&space;Flight,&space;t&space;=\frac{2v_{0sin\theta}}{g}$

$\large&space;Maximum\;&space;height\;&space;reached,&space;H&space;=\frac{v^{2_{0sin^{2\theta}}}}{2g}$

$\large&space;Horizontal&space;range,R=\frac{v^{2_{0}sin2\theta}}{g}$

Where,
the initial Velocity is Vo ,
the component along the y-axis is sin θ,
the component along the x-axis is cos θ.

Projectile Motion formula is made use of to calculate the distance, velocity and time engaged in the projectile motion.

Projectile Motion Solved Examples

Underneath are questions based on projectile motion which may help one in their exam.

Problem 1: A body is projected with a velocity of 20 ms-1 at 50o to the horizontal. Find
(i) Maximum height reached
(ii) Time of flight and
(iii) Range of the projectile.

Initial Velocity Vo = 20 ms-1,

$\large&space;\theta&space;=&space;50^{\circ}$

$\large&space;Time&space;of&space;flight,t=&space;\frac{2V_{0}sin\theta}{g}$

$\large&space;=\frac{2\times20\times&space;sin50^{\circ}}{9.8}$

= 3.126s

Maximum height reached, H = $\large&space;\frac{V^{2}sin^{2}\theta}{2g}$

$\large&space;\frac{(20)^{2}sin^{2}50^{2}}{2\times9.8}$

=11.97m

Horizontal  Range R = $\large&space;\frac{v^{2}_{0}sin2\theta}{g}$

$\large&space;=\frac{20^{2}sin100\theta&space;}{9.8}$

=40.196m

Problem 2: John is on top of the building, and jack is down. If John tosses a ball at an angle of 60o and with an initial velocity of 20 m/s, what height will the ball touch after 2 s?