A ball is thrown at a speed of 15m/s at a 30 degree angle from the floor, how far does this ball travel before hitting the ground? (Air resistance can be ignored)

The key to answering this question is to separate the velocity of the ball into two components - horizontal and vertical components. The vertical component will be affected by the gravitational acceleration (take this value to be 9.81m/s²; the horizontal component will remain the same for the duration of flight as the question states air resistance can be ignored. By drawing a diagram and using trigonometry, the vertical velocity component is found to be 15sin(30) m/s. This doesn't affect the horizontal distance travelled, but the time of flight can be found from this by using the SUVAT equation v = u+at. It's important to set the final velocity as -15sin(30) m/s (due to energy conservation, the velocity of the ball just before hitting the ground will be the same as the initial vertical velocity but in the opposite direction). Solving the equation for time, we find that t = 1.53s
Using this, the total distance travelled can be found by multiplying HORIZONTAL (using vertical velocity is a common mistake) velocity with the time of travel. 15*cos(30)*1.53= 19.86 meters. Final answer = 19.9 meters (rounded to 3 s.f.)