When a Golf player hits a Golf ball, a baseball player hits a baseball or a tennis player hits a tennis ball then you can notice that the motion of these balls are of the same kind and it is called a Projectile motion.
What is Projectile Motion?
When you throw an object at an angle, then the motion of that object is called projectile motion.
Suppose a ball is kicked at an angle. Assuming the medium is a vacuum. Let’s choose a coordinate system such that an x-y plane is a vertical plane as shown in the animation. Here only one force is acting on the ball and this is the gravitational force that is constant and acting along the -y-axis. So, the acceleration of the ball along the x-axis is a_x = 0 and the acceleration along the z-axis is a_y = -g = -9.8\rm \ m/s^2. Play the video to see how the velocity of the ball is changing.
Let the ball leaves the ground at the time, t = 0 with the velocity v_0 along a direction that makes an angle \theta with the x-axis. Also assume that the total time taken by the ball to reach the ground is T, the range is R, and the maximum height reached by the ball is H as shown in the above animation.
- At x = 0 and t = 0
v_x = v_0 \cos \theta and v_y = v_0 \sin\theta.
- At x = x_1,
v_x = v_0\cos\theta and v_y < v_0\sin\theta
- At x = \frac{R}{2}, t = \frac{T}{2} and y = H
v_x = v_0 \cos \theta and v_y = 0
- At x = x_2,
v_x = v_0\cos\theta and v_y < v_0\sin\theta .
- At x = R, t = T and y = 0,
v_x = v_0 \cos \theta and v_y = v_0 \sin\theta.
(Since acceleration is only along the -y-axis so the velocity along the y-axis is decreasing. As acceleration along the x-axis is zero so the velocity along the x-axis will be constant throughout the motion of the ball.)
Now in our case, the acceleration is constant so, we can apply the equations of kinematics to find the equation of the trajectory, the time of flight, the maximum height, and the range of the projectile motion.
We have four equations of kinematics. Those equations are,
v = v_0 + at
\textcolor{red}{(1)}
\Delta S =\dfrac{v+v_0}{2} \times t
\textcolor{red}{(2)}
\Delta S = v_0t+\dfrac{1}{2}at^2
\textcolor{red}{(3)}
v^2 = v_0^2 + 2a \Delta S
\textcolor{red}{(4)}
Equation of Trajectory
At any time t, if you consider the motion of the ball only along the y-axis then, by using the equation (3), the y-coordinate of the ball is given by;
\begin{aligned}y &= (v_0)_y t + \dfrac{1}{2} a_y t^2\\&= v_0 \sin\theta t + \dfrac{1}{2} (-g) t^2\\&= v_0 \sin\theta t – \dfrac{1}{2} g t^2 \\\end{aligned}
\textcolor{red}{(5)}
If you consider the motion of the ball only along the x-axis then, by using equation 3, the x-coordinate of the ball is given by;
\begin{aligned}x &= (v_0)_x t + \dfrac{1}{2} a_x t^2\\&= v_0 \cos\theta t + \dfrac{1}{2} \times 0 \times t^2\\&= v_0 \cos\theta t\\\end{aligned}
\textcolor{red}{(6)}
If you plug the value of t from equation 6 into equation 5, then equation 5 becomes
\begin{aligned}y &= v_0 \sin\theta (\dfrac{x}{v_0 \cos\theta}) – \dfrac{1}{2} g (\dfrac{x}{v_0 \cos\theta})^2\\\end{aligned}
And finally, it can be written as;
\fcolorbox{black}{aqua}{$y = x tan\theta – \dfrac{gx^2}{2v_0^2 \cos^2\theta}$}
\textcolor{red}{(7)}
Equation 7 is called the equation of the trajectory of the projectile motion.
Time of Flight
Considering the motion of the ball along the y-axis during the time t = 0 to t = T/2.
We have,
- the initial velocity is (v_0)_y = v_0 \sin\theta,
- the final velocity is v_y = 0,
- the acceleration is a_y = -g,
- the displacement is \Delta S_y = H and
- the time interval is t = \dfrac{T}{2}
By using equation 1, you can write
\begin{aligned}v_y &= (v_0)_y + a_y t\\\Rightarrow 0 &= v_0 \sin\theta -g \times \dfrac{T}{2}\\\end{aligned}
And we get
\fcolorbox{black}{aqua}{$T = \dfrac{2v_0 \sin\theta}{g}$}
\textcolor{red}{(8)}
This is the formula for the time of flight.
Maximum Height
By using equation 4, you can write
\begin{aligned}v_y^2 &= (v_0)_y^2 + 2a_y \Delta S_z\\\Rightarrow 0^2 &= (v_0 \sin\theta)^2 +2(-g) H\\\end{aligned}
And we get
\fcolorbox{black}{aqua}{$H = \dfrac{v_0^2 \sin^2\theta}{2g}$}
\textcolor{red}{(9)}
This is the formula for the maximum height.
Range
Considering the motion of the ball along the x-axis during the time t = 0 to t = T.
We have,
- the initial velocity is (v_0)_x = v_0 \cos\theta,
- the acceleration is a_x = 0,
- the displacement is \Delta S_x = R and
- the time interval is t = T.
By using equation 3, you can write
\begin{aligned}\Delta S_x &= (v_0)_x t + \dfrac{1}{2} a_x t^2\\\Rightarrow R &= v_0 \cos\theta T + \dfrac{1}{2} \times 0 \times T^2\\&= (v_0 \cos\theta)\times \dfrac{2v_0 \sin\theta}{g} + 0\\&= \dfrac{v_0^2 (2sin\theta \cos\theta)}{g}\\\end{aligned}
Since 2sin\theta \cos\theta = \sin2\theta so we get
\fcolorbox{black}{aqua}{$R = \dfrac{v_0^2 \sin2\theta}{g}$}
\textcolor{red}{(10)}
This is the formula for the Range.
These formulas are valid only for the case when the point at which the projectile is launched and the point where it hits the ground is at the same horizontal level. For all other cases, these formulas are not valid, So, you must know how to use the equations of kinematics, and by doing that you can solve the problems of any type of projectile motion.