Question

Four identical balls are thrown from the top of a cliff, each with the same speed. The

Answer 1

Answer:

the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

Explanation:

The kinetic energy, K.E. = (1/2) × m × v²

The velocity of the ball, v = u × sin(θ)

Where;

u = The initial velocity of the ball

θ = The reference angle

1) For the ball thrown straight up, we have;

θ = 90°

∴ v = u

The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh

Where;

h = The height of the cliff

∴ K.E. = (1/2) × m × (u² + 2gh)²

2) For the second ball thrown 30° to the horizontal, we have;

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

3) For the third ball thrown at 30° below the horizontal, we have;

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

4)  For the fourth ball thrown straight down, we have;

K.E. = (1/2) × m × (u² + 2gh)²

Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

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