Final answer:
The hopper leaves the floor with a velocity of approximately 5.05 m/s directed upwards.
Step-by-step explanation:
To find the velocity at which the hopper leaves the floor, we can use the principles of projectile motion. Given that the hopper jumps straight up to a height of 1.3 m, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which is 0 because the hopper starts from rest), a is the acceleration (which is -9.8 m/s^2 due to gravity), and s is the distance traveled (which is 1.3 m).
Using the equation, we can solve for v:
v^2 = 0^2 + 2(-9.8)(1.3)
v^2 = -25.48
v = -5.05 m/s
Since velocity cannot be negative in this context, we can conclude that the hopper leaves the floor with a velocity of approximately 5.05 m/s directed upwards.