Supposing the runner is condensed to a point and moves upward at 2.2 m/s.
It takes a time = 2.2/g = 2.2/9.8 = 0.22 seconds to increase to max height.
Now looking at this condition in opposite - that is the runner is at max height and drops back to earth in 0.22 s (symmetry of this kind of motion).
From what height does any object take 0.22 s to fall to earth (supposing there is no air friction)?
d = 1/2gt²= (0.5)(9.8)(0.22)²= 0.24 m
Answer:
6m/s
Explanation:
Given parameters:
Initial velocity = 0m/s
Acceleration = 2m/s²
Distance = 9m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we use the expression below:
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
v² = 0² + (2 x 2 x 9) = 36
v = 6m/s
