As a 5.0 × 10^2-newton basketball player jumps
from the floor up toward the basket, the
magnitude of the force of her feet on the floor is
1.0 × 10^3 newtons. as she jumps, the magnitude
of the force of the floor on her feet is
(1) 5.0 × 10^2 n (3) 1.5 × 10^3 n
(2) 1.0 × 10^3 n (4) 5.0 × 10^5 n

     This player is initially at rest, then the Force Weight (W) and the Normal Force (N) have a same module. When applies a Extra Force (E) on the floor, this reacts with this increase, causing the player to be released upwards.
     Using the Newton's Secound Law, we have:



Number 3

If you notice any mistake in my english, please let me know, because i am not native.

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way