Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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

share|improve this answer
    
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. –  in_wolframAlpha_we_trust Apr 17 '12 at 8:35
    
Sorry, didn't notice it then. It's zero at the mid-point I guess, so just halved the value accordingly. –  Wonder Apr 17 '12 at 8:38
    
Yeah, that looks good. –  in_wolframAlpha_we_trust Apr 17 '12 at 9:55
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.