Ahopper jumps straight up to a height of 1.3 m. with what velocity did he leave the floor

5.047m/s

Explanation:

Initial velocity (U) = ?

Height (s) = 1.3m

Time taken (t) =?

g = acceleration due to gravity = 9.8m/s²

From equation of motion;

S = ut - ½gt²equation (i) (note the negative sign is because the Hooper is moving against gravity).

Velocity at maximum height = 0

V = u - gt equation (ii)

0 = u - gt

t = u/g

Put t = u/g into equation (i)

S = u(u/g) - ½g(u/g)²

S = u²/g - ½u²/g

S = ½u²/g

2S = u²/g

U² = 2Sg

U² = 2 * 1.3 * 9.8

U² = 25.48

U = 5.047m/s (take square root of U²)