A cannonball is fired at an angle of 30 degrees and a velocity of 16 m/s. How long does it take (to 2 significant figures) for the cannonball to reach the ground?

First, we make the standard assumption that there is no air resistance, meaning that gravity is the only force acting on our cannonball. We then use the SUVAT equations, so: S = 0m (because it will return to the ground), U = -16sin30 (because we look at the vertical velocity) m/s, V is unknown, A = g = 9.8 m/s^2 and T=t. Then, we can use the kinematics equation s = ut + 0.5t^2.Substituting our values, 0 = -16sin30(t) + 0.5x9.8t^2. Then, using the quadratic formula, we get two solutions for t: t = 1.63s and t = 0s. We eliminate the 0 solution because that is the launch position. Thus it takes 1.63s to 2s.f.