Physics > Motion in Two Dimension > 3.0 Ground to ground projectile motion
Motion in Two Dimension
1.0 Introduction
2.0 Projectile motion
3.0 Ground to ground projectile motion
3.1 Maximum height
3.2 Time of flight
3.3 Range
3.4 Trajectory of a projectile
3.5 Summary
3.6 Solved Examples
4.0 Projectile thrown parallel to the horizontal
5.0 Projectile on an inclined plane
6.0 Relative motion between two projectiles
3.2 Time of flight
3.2 Time of flight
3.3 Range
3.4 Trajectory of a projectile
3.5 Summary
3.6 Solved Examples
${\overrightarrow u _y} = u\sin \theta $
${\overrightarrow a _y}=-g$
${\overrightarrow s _{y}}=0$
$t=T$
Since the particle returns back to the ground. So, the vertical displacement is $0$.
Therefore we can write kinematics equation which relates displacement and time as,
$${\overrightarrow s _y} = {\overrightarrow u _y}t + \frac{1}{2}{\overrightarrow a _y}{t^2}$$$$0 = u\sin \theta \ T + \frac{1}{2}\left( { - g} \right){T^2}$$$$T = \frac{{2u\sin \theta }}{g}$$