A projectile is thrown from the ground at 30 degrees from the horizontal direction with an initial speed of 20m/s. What is the horizontal distance travelled before it hits the ground? Take the acceleration due to gravity as 9.8m/s^2

Draw diagram outlining the symmetric parabolic shape of the projectile's motion. Find vertical component of the initial speed using SOH CAH TOA. sin(30) = opposite/hypotenuse = x/30therefore, x = 30sin(30)vertical acceleration is -9.8 (negative because taking positive motion as vertically upwards)at the top of the parabolic curve, the projectile will have zero vertical velocity, therefore take final vertical velocity = 0using suvat equation: v^2 = u^2 + 2as with v=0, u=30sin(30) and a=-9.8subbing in and solving for s gives s=11.5mThis is the distance for the projectile to reach the maximum height in its path. The curve is symmetric so to find the total distance before it hits the floor, double this value giving the horizontal distance covered=23m