A sprinter is just coming out of the starting block, and only one foot is touching the block. The sprinter pushes back horizontally against the block with a force of 800 N and pushes down vertically against the ground with a force of 1000 N. How large is the resultant force and in what direction and angle is the resultant force located

1280.63 N, 231.35°

Explanation: Both the vertical and horizontal force are perpendicular to each other.

The horizontal force is 800 N to the back ( which corresponds to a force of 800 N acting in the negative x direction).

The vertical force is 1000 N acting down vertically against the ground ( which corresponds to a force of 1000 N acting in the negative y direction)

Hence the resultant force is placed in the third quadrant ( between the negative x and negative y axis).

The magnitude of resultant force is gotten below as

R = √(Fx)² + (Fy)2

Fx = 800 and Fy = 1000 N

R = √800² + 1000²

R = √640000 + 1000000

R =√1640000 = 1280.63 N

direction = tan^-1 ( 1000/800) = tan^-1 (1.25) = 51. 35°

But the force is on the third quadrant, hence angle is 180 + 51.35° = 231.35°

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