# Force required to jump

If I wanted to know how much force is required to jump, and I know my acceleration while jumping, would I just use F=ma. Or do I need to account for gravity, as in use
Fjump - Fg = ma

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if the acc. you find is the net acc. then i guess you donot need to consider gravity as ots already considered while finding the net acc.

Yeah, thats what I was thinking.
Initially I was given the final velocity when jumping and the time in contact with the ground.
Vf = vi + at
That should be my net acceleration, because that is the actual acceleration needed to reach Vf.

If the question is what force do I need to push off the ground with, then aren't I looking for Fnormal and not net force?
The force that I push off the ground with is equal/opposite to the force normal acting on me.
So shouldn't I be solving FN - Fg = ma for FN?

please can you provide full question

Question:
If someone is able to jump 2.0 m into the air
a) How what speed must they leave the ground with?(ignore air resistance)
b) If they are in contact with the ground for 0.25 sec and have a mass of 65kg, what force must they push off the ground with?

Solution:
a) Going from the ground to max height
vi = ?, vf = 0m/s, a = -9.8m/s2, $$\Delta$$y = 2.0m
vf2 = vi2+2ay
vi = 6.26m/s

b) just for the jumping part
vi = 0m/s, vf = 6.26m/s, a = ?, t = 0.25
Find acceleration using vf=vi +at
a = 25.04 m;s2

Now, as you read the question, do you think it's asking for the net force applied to the person to jump or is it asking what the person must push off with?

net force acting on man must be now ma
this will include the gravity force and the normal reaction acting on him

Now, as you read the question, do you think it's asking for the net force applied to the person to jump or is it asking what the person must push off with?
I guess they are asking the force with which the man must push the ground, which is the normal rxn acting on it

I guess they are asking the force with which the man must push the ground, which is the normal rxn acting on it
Yes I agree with you. I think they are asking for the normal force.
Thanks for the help.