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Projectile Motion


When an object is launched in the air at an angle, its motion is two-dimensional and if it is only affected by gravity, this motion is known as Projectile Motion.






Image illustrating Projectile Motion





  • If we are given a velocity vvwith angle of θ\thetarespect to horizontal direction:


vx0=v0 cos⁡θv_{x0}=v_0\ \cos\theta


vy0=v0 sin⁡θv_{y0}=v_0\ \sin\theta


  • We can find time of flight to maximum height, maximum height and range of motion using the following equations respectively:


t=v0sinθgt = \dfrac{v_{0}sin\theta}{g}


h=v02sin2θ2gh = \dfrac{v^{2}_{0}sin^{2}\theta}{2g}


R=v02sin2θgR = \dfrac{v^{2}_{0}sin2\theta}{g}







Some Useful Notes about Projectile Motions

− Find the time to reach the ground from the y-component calculations.

− Find the horizontal distance (or range) from the x-component calculations.

− The velocity along the y-axis changes. vyo=0v_{yo}=0 at maximum height.

− Projectile motion is completely symmetric. For example, time to go up from a certain height equals time to come back down to that same height.

− Maximum range (horizontal distance) is obtained when the launch angle is 45°.




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Example: A Ball in Projectile Motion


A ball is shot at an angle of 30° from the horizontal. If the ball had an initial speed of 20m/s, how long was it in flight before it hit the ground?



Image illustrating Example: Projectile Motion

In y-direction:

vo=20m/sLets calculate the time ball takes to reach the maximum heightv0y=20sin(30)=10m/say=−9.8m/s2t=?vfy=voy+ayt0=10+(9.8)tt=−10−9.8=1.02 sec    total time during the flight is T=2t=2sec\begin{array}{ll} v_o=20m/s\\ \text{Lets calculate the time ball takes to reach the maximum height}\\ v_{0y}=20sin(30)=10m/s\\ a_y=-9.8m/s^2\\ t=?\\ v_{fy}=v_{oy}+a_yt\\ 0=10+(9.8)t\\[3pt] t=\frac{-10}{-9.8}=1.02~sec\ \ \ \text{ total time during the flight is }\\ T = 2t = 2 sec \end{array}



A basketball player shoots a ball from 2-m height above the ground at angle of 30° respect to the horizontal direction toward a net. He scores in the net, which is 4.5 meters above the ground and 7.5 meters away from where the player stands.

What is the speed of the ball at which it was shot?

Choose an answer

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Extra Practice