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A 265-gram ball is thrown at a speed of 35.2 m/s from the top of a 39.1-meter high cliff. Determine the impact speed of the ball when it strikes the ground. Assume negligible air resistance.

Answer :

A 265-gram ball is thrown from the top of a 39.1-m high cliff with an initial velocity of 35.2 m/s. Neglecting air resistance, the impact speed when the ball strikes the ground is approximately 44.80 m/s in the upward direction.

To determine the impact speed of the ball when it strikes the ground, we can use the principles of kinematics and conservation of energy. We'll consider the initial vertical displacement (Δy), initial vertical velocity (Viy), and final vertical velocity (Vfy) of the ball.

Mass of the ball (m) = 265 grams = 0.265 kg

Initial velocity (Viy) = 35.2 m/s (upward direction, since the ball is thrown)

Height of the cliff (h) = 39.1 m

Acceleration due to gravity (g) = 9.81 m/s²

Calculate the initial vertical displacement (Δy):

Δy = h = 39.1 m (as the ball is thrown from the top of the cliff)

Calculate the final vertical velocity (Vfy) using the kinematic equation:

Vfy² = Viy² + 2 * g * Δy

Vfy² = (35.2 m/s)² + 2 * (9.81 m/s²) * (39.1 m)

Vfy² = 1239.04 m²/s² + 766.722 m²/s²

Vfy² ≈ 2005.762 m²/s²

Vfy ≈ √(2005.762 m²/s²)

Vfy ≈ 44.80 m/s (upward velocity at the ground level)

Calculate the impact speed of the ball when it strikes the ground:

The impact speed of the ball is the magnitude of the final vertical velocity (Vfy) at the ground level since it's moving upward.

Impact speed ≈ 44.80 m/s

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