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

• Silverbolt
In summary, to calculate the speed needed to jump a distance of 3.7m at an angle of 25° in the long jump event, we can use the equation vf²=vi²+2aY (final velocity squared equals initial velocity squared plus 2 times acceleration times distance in the y-direction). Plugging in the values, we get a final velocity of 13.1m/s. However, it should be noted that this equation is for vertical motion, not horizontal motion as in the long jump. In projectile motion, the time of flight can be found by using the vertical component of the velocity and the range can be found by using the horizontal component of the velocity.
Silverbolt
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

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

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.