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

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)

Answered by aabdulsamir | 2024-06-12

The hopper leaves the floor with a velocity of approximately 5.05 m/s directed upwards. ;

Answered by qwviolet | 2024-06-18

The hopper leaves the floor at a velocity of approximately 5.05 m/s to reach a height of 1.3 m. This calculation is derived from applying the kinematic equation relating initial velocity, final velocity, acceleration, and displacement. Using the acceleration due to gravity, we rearranged the equation to isolate and solve for initial velocity.
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Answered by aabdulsamir | 2024-09-26