I need help for the following question:
In a long jump event, an athlete jumps a horizontal distance of 7.56m. The athlete was airborne for 3.04 seconds. The acceleration due to gravity is taken as 9.81m/s^2. Assume that air resistance is negligible, calculate his take-off speed (initial velocity) in m/s.
Any comments and feedback on how to tackle the question will be much appreciated.
Best Answer
Let initial velocity v have horizontal component v1 m/s and vertical component v2 m/s. v2 = 9.81 * 3.04/2 = 14.9112. v1 = 7.56/3.04 = 2.48684211. Speed I guess will be norm of v = $\sqrt{v1^2 + v2^2} = \sqrt{2.48684211^2 + 14.9112^2} = 15.1171515 m/s$.