Question

A hopper jumps straight up to a height of 1.3 m. With what velocity did he leave the floor

Answer 1

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.

Answer 2

Answer: 5.047m/s

Step-by-step 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 - ½gt².........equation (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)