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 specific speed when launched at a 25° angle. The initial discussion suggested using the equation vf²=vi²+2aY, but this is not applicable for horizontal distance in projectile motion. Instead, the time of flight should be calculated using the vertical component of the velocity, while the horizontal range is determined by the horizontal component. The correct approach involves analyzing both components separately to find the required speed for the jump. Understanding the distinction between vertical and horizontal motion is crucial for solving this problem accurately.
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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