AP Physics ~ A good jump in the long jump event is

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A good jump in the long jump event is 3.7m, requiring a takeoff speed of 13.1m/s at a 25° angle. The calculation uses the equation vf²=vi²+2aY, where the final vertical velocity is zero at the peak of the jump. The jumper's initial vertical velocity is determined by the sine of the angle and the necessary distance. Participants are reminded to post homework questions in the designated forum section for better organization. Proper forum etiquette enhances the discussion experience.
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A good jump in the long jump event is 3.7m. If the jumper leaves the ground at an angle of 25°, what speed must he need to jump the 3.70m distance?

→Could you use this equation: vf²=vi²+2aY aka (final velocity)²= (initial velocity)²+2(acceleration)(distance in the y-direction)
→So, that would be: 0=(vsin25 °)²+2(-9.8)(3.7)∴ v=13.1m/s

Please help me if I am wrong.
 
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