Projectile & Projectile Motion – definitions
In this post we would define and discuss on Projectile and Projectile Motion. We will also derive Projectile Motion Equations and find out how the motion path looks like under the influence of both horizontal and vertical component of the projectile velocity.
When we throw some object making an angle with the horizontal (angle less than 90 degree), then what is/are the force/s acting on that object? As I throw a stone like this and it goes off my hand then I am no more applying any force on that object. But there will be a force omnipresent acting on that stone. That is Gravity or Gravitational force applied by the earth. And this force directs vertically downwards pointing to the centre of the earth. If gravity was not present that stone would continue to move in a straight line maintaining the same angle with horizontal plane. But this downward pull of gravity takes it down. In one hand that stone tries to maintain a constant velocity along the horizontal as there is no force on it along that direction. On the other hand it gets a downward pull caused by gravity and faces an acceleration equal to g (acceleration due to gravity). Under these 2 movement it takes a specific path, which we will discuss now.
Projectile definition: When an object is in flight after being projected or thrown then that object is called a projectile and this motion is called Projectile Motion. Example, the motions of cricket ball, base ball.
Projectile motion definition: The motion of a projectile is known as projectile motion which is basically result of 2 separate simultaneously occurring components of motions. One component is along a horizontal direction without any acceleration (as no force acting in this direction) and the other along the vertical direction with constant acceleration due to the force of gravity.(considering air resistance as nil)
Projectile Motion Equations: In the next sections we will discuss and derive couple of projectile motion equations.
Projectile Motion Equations
We will cover here couple of equations like the
- the projectile path equation
- equation for time to reach maximum height
- total time
- maximum height of a projectile and
- equation for horizontal range of a projectile
Projectile Motion equation derivation – parabola
Say an object is thrown with uniform velocity V0 making an angle theta with the horizontal (X) axis.
The velocity component along X axis = V0x = V0 cosθ and the velocity component along Y axis = V0y = V0sinθ. Air resistance is taken as negligible.
At time T =0 , there is no displacement along X and Y axes. So X0=0 and Y0=0
At time T=t,
Displacement along X axis = x= V0x.t = (V0 cosθ). t …………….. (1) and
Displacement along Y axis = y= V0y.t = (V0sinθ ).t – (1/2) g t2 …………(2)
From equation 1 we get: t = x/(V0 cosθ) ……………….. (3)
Replacing t in equation 2 with the expression of t from equation 3:
y = (V0sinθ ). x/(V0 cosθ) – (1/2) g [ x/(V0 cosθ)]2 = (tanθ) x – (1/2) g . x2/(V0 cosθ)2………………….. (4)
In the above equation g, θ and V0 are constant.
So rewriting equation 4:
y = ax + bx2 where a and b are constants.
This is an equation representing parabola.
So we can say that the motion path of a projectile is a parabola. So once thrown a projectile will follow a curved path named as Parabola.
Time to reach the Maximum Height by a projectile
When the projectile reaches the maximum height then the velocity component along Y axis i.e. Vy becomes 0. Say the time required to reach this maximum height is tmax. The initial velocity for the motion along Y axis (as said above) is V0sinθ
Considering vertical motion along y axis:
Vy = V0sinθ – g tmax
=> 0 = V0sinθ – g tmax
=> tmax= (V0sinθ )/g ……………………………..(5)
So this is the equation for the time required to reach the maximum height by the projectile.
Total time of flight for a projectile:
So to reach the maximum height by the projectile the time taken is (V0sinθ )/g
It can be proved that the projectile takes equal time [ (V0sinθ )/g] to come back to ground from its maximum height.
Therefore the total time of flight for a projectile Tmax = 2(V0sinθ )/g …………………. (6)
Maximum Height reached by a projectile
Let’s say, the maximum height reached is Hmax .We know that when the projectile reaches the maximum height then the velocity component along Y axis i.e. Vy becomes 0.
Using one of the motion equations, we can write
(Vy)2 =( V0sinθ )2 – 2 g Hmax
=> 0 = ( V0sinθ )2 – 2 g Hmax
=> Hmax = ( V0sinθ )2/(2 g) …………………… (7)
Horizontal range of a projectile
Say R is the horizontal range covered by a projectile is R when it touches the ground after passing the time Tmax
Here we will use the horizontal component of the velocity only as the effective velocity to traverse this horizontal path.
R = (V0 cosθ) . Tmax = (V0 cosθ). 2(V0sinθ )/g = (V02 sin2θ )/ g
R = (V02 sin2θ )/ g ………………………. (8)
For Maximum value of R , it is pretty easy to understand from the above equation that the angle θ needs to be equal 45 degree, because in that case we get the maximum value from sine as 2θ becomes 90 degree.
Hence Rmax = V02 / g …………………………. (9)
Related study: Equations for free fall for vertical motion