Force applied to the ground from a 1.2 m standing vertical jump

[SteveVer256]

Okay, here's the deal:

Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 m (4 ft). (This means that he moved upward by 1.2 m after his feet left the floor.) Griffith weighed 890 N (200 lb).

(B) If the time of the part of the jump before his feet left the floor was 0.300 s, what was the magnitude of his average acceleration while he was pushing against the floor?
16.2 m/s Correct

Use Newton's laws and the results of part (B) to calculate the average force he applied to the ground.

Okay, I tried various methods of solving this that have all failed (says it is wrong). I have two tries left - so any ideas?

Numbers I have tried:
1480
1470
291
1180

If I am close to within about 5 digits, it will tell me. So far, I haven't even had that luck.

Anyone?
__________________

 

[krab]

16.2 m/s cannot be correct, as it is a velocity. The question asks for acceleration. I get 16 feet per second squared.

 

[Doc Al]

(B) If the time of the part of the jump before his feet left the floor was 0.300 s, what was the magnitude of his average acceleration while he was pushing against the floor?
16.2 m/s Correct

Right. I assume you meant 16.2 $m/s^2$. (Units matter!)

Use Newton's laws and the results of part (B) to calculate the average force he applied to the ground.

Identify the forces acting on the man, then apply Newton's 2nd law:
$$F_{net} = ma$$.

Realize, from Newton's 3rd law, that the force that the floor exerts on the man is equal and opposite to the force that the man exerts on the floor.