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

Silverbolt
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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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Silverbolt said:
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.

It is in x direction(long jump) not y direction(high jump)
 
You can't use the formula which you have used.
In the projectile motion, the time of flight can be found by using vertical component of the velocity and the range can be found by using horizontal component of the velocity. Now try to solve the problem.
 

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