A golf ball is hit from horizontal ground with speed 10 m/s at an angle of p degrees above the horizontal. The greatest height the golf ball reached above ground level is 1.22m. Model the golf ball as a particle and ignore air resistance. Find p.

Initial horizontal speed of particle = 10cos(p) m/s. Initial vertical speed of particle = 10sin(p) m/s. ('U' in suvat.) There are no forces other than gravity acting on the particle so the vertical acceleration on the partical while it is moving upwars is -9.8 m/s². ('A' in suvat.) The greatest height reached by the golf ball is 1.22m. ('S' in suvat.) At this point, the ball has a vertical velocity of 0 m/s ('V' in suvat) as it is not moving upwards or downwards. Using this information, obtained from the question, we find out p using the suvat equation V² = U²+2AS. 0² = (10sin(p))² +2(-9.8)(1.22) 100sin² (p) -23.912=0 sin²(p) =0.23912 sin(p)=0.4889989... p=sin^-1(0.488989...). p=29.3.