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The Bridge and the Rock Using energy considerations and assuming negligible air resistance, find the speed of a rock just before it hits the water if it is thrown from a bridge 16.4 m above water with an initial speed of 16.4 m/s? (Verify that the speed is independent of the direction the rock is thrown.)

User Tariq Aziz
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2 Answers

1 vote

Final answer:

The final speed of the rock just before it hits the water can be found using the conservation of energy, yielding approximately 24.3 m/s. This speed is independent of the direction of the throw due to energy considerations.

Step-by-step explanation:

Calculating the Speed of a Rock Thrown from a Bridge

To find the speed of the rock just before it hits the water, we will use the principle of conservation of energy. This physics principle states that the total mechanical energy (kinetic plus potential) in a system remains constant if only conservative forces (like gravity) are doing work.

Initially, when the rock is thrown from the bridge, it has both kinetic energy (due to its initial speed) and gravitational potential energy (due to its height above the water). As it falls, potential energy is converted into kinetic energy.

The initial kinetic energy (Ki) can be given as:

Ki = 0.5 * m * vi2

The potential energy (Ui) at the height h is:

Ui = m * g * h

Just before the rock hits the water, the potential energy is zero and all the energy is kinetic. The final kinetic energy (Kf) is therefore equal to the sum of the initial kinetic and potential energies:

Kf = Ki + Ui

0.5 * m * vf2 = 0.5 * m * vi2 + m * g * h

When we solve for vf, the mass (m) cancels out, and we get:

vf = sqrt(vi2 + 2 * g * h)

Substituting the given values, we have:

vf = sqrt(16.42 + 2 * 9.8 * 16.4)

vf = sqrt(268.96 + 321.44)

vf = sqrt(590.4)

vf = 24.3 m/s (approximately)

This calculation shows that the final speed of the rock just before it hits the water is approximately 24.3 m/s, and it is indeed independent of the direction the rock was thrown.

User Karl L
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3 votes

Answer:

24.30 m/s

Step-by-step explanation:

If the stone is thrown horizontally, our horizontal velocity u₁ = 16.4 m/s and its initial vertical velocity, U = 0 and its final vertical velocity, V is thus

V² = u² + 2gs where s = height of bridge above water = 16.4 m, g = 9.8 m/s²

v² = 0 + 2 × 9.8 × 16.4 = 321.44

v = √321.44 = 17.93 m/s

Since the horizontal velocity is constant, our resultant velocity just before it hits the water is v₁ = √(u₁² + v²) = √(16.4² + 17.93²) = √(268.96 + 321.44) = √590.4 = 24.30 m/s

If the stone is thrown vertically, its initial vertical velocity, u = 16.4 m/s. its final vertical velocity, v is thus

V² = u² + 2gs where s = height of bridge above water = 16.4 m

v² = 16.4² + 2 × 9.8 × 16.4 = √(268.96 + 321.44)

v = √590.4 = 24.30 m/s

User S Gaber
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