A projectile is fired at a speed of 10m/s at an angle of 30 degrees to the horizontal. What is the time of flight of the projectile and what is its range?

Projectile motion can be split into its horizontal and vertical components. The horizontal component of the initial velocity will remain constant as no force acts in the horizontal direction. The vertical component experiences constant acceleration due to gravity. In this problem the vertical component of velocity is equal to 10sin(30) = 5m/s and the horizontal component is 10cos(30) = 5sqrt(3).We can use the suvat equations for motion in the vertical component as gravity provides constant acceleration. The time taken to reach the top of the flightpath is given by 0 = 5-gt therefore t = 5/g. For the total time of flight we double this to give t = 10/g seconds. To calculate the range of the projectile we mulitply the horizontal velocity by the time of flight 5sqrt(3) x 10/g = (50sqrt(3))/g metres .