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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.

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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 = $\sqrt{v1^2 + v2^2} = \sqrt{2.48684211^2 + 14.9112^2} = 15.1171515 m/s$.

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  • $\begingroup$ I don't think you're calculating the vertical component of his speed correctly. The change in his speed is given by g*t but his vertical velocity is not zero at take-off or at landing. $\endgroup$ – in_mathematica_we_trust Apr 17 '12 at 8:35
  • $\begingroup$ Sorry, didn't notice it then. It's zero at the mid-point I guess, so just halved the value accordingly. $\endgroup$ – Wonder Apr 17 '12 at 8:38
  • $\begingroup$ Yeah, that looks good. $\endgroup$ – in_mathematica_we_trust Apr 17 '12 at 9:55
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You need to calculate the horizontal initial velocity and vertical initial velocity separately. For the vertical part, you know that the time the athlete traveled and the height he reached. For the horizontal part, just note that since air resistance is neglected, the initial horizontal velocity will remain the same throughout the process.

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