A ball is kicked and has an instantaneous velocity of 19.6m/s at an angle of 30 degrees to the horizontal. A target lies flat on the ground in the direction the ball is kicked and lies at a distance of (98/5)*(3^1/2)m. Does the ball land on the target?

First assume air resistance is negligible.Draw a diagram for the problem, this makes picturing it a lot easier.Next calculate the horizontal and vertical velocity of the ball. With the aid of a diagram and trigonometry (SOHCAHTOA) the vertical speed is found to be 19.6sin(30) and the horizontal 19.6cos(30).Now use SUVAT to find the time that the ball is in the air.At the top of the arc the ball makes, it has a vertical velocity of 0. Therefore I can use v=u+at to find my time from launch to peak. This gives t = 1s (rearranging gives t = u/a, t = 19.6sin(30) / 9.8) Due to symmetry this can be doubled to give the total time the ball is airbourne. T(total) = 2s.Using this we can calculate the horizontal distance the ball travels whilst in the air. Use the equation distance = speed x time.Time = 2s, horizontal speed = 19.6cos(30). Therefore distance = 19.6*cos(30)*2. using a calculator or a 'magic' triangle this is seen to be equivalent to (98/5)*3^1/2.Yes, the ball hits the target.