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## Main Question or Discussion Point

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.

→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.