- #1
- 24
- 0
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.