Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of what, independent of the direction thrown.

Complete question is;

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of 31.1 m/s, independent of the direction thrown

It is proved that the final speed is truly 31.1 m/s

Explanation:

From energy - conservation principle;

E_i = Initial potential energy + Initial Kinetic Energy

Or

E_i = U_i + K_i

Similarly, for final energy

E_f = U_f + K_f

So, expressing the formulas for potential and kinetic energies, we now have;

E_i = (m × g × y_i) + (½ × m × v_i²)

Similarly,

E_f = (m × g × y_f) + (½ × m × v_f²)

We are given;

y_i = 25 m

y_f = 0 m

v_i = 20 m/s

v_f = 31.1 m/s

So, plugging in relevant values;

E_i = m((9.8 × 25) + (½ × 20²))

E_i = 485m

Similarly;

E_f = m((9.8 × 0) + (½ × v_f²)

E_f ≈ ½m•v_f²

From energy conservation principle, E_i = E_f.

Thus;

485m = ½m•v_f²

m will cancel out to give;

½v_f² = 485

v_f² = 485 × 2

v_f² = 970

v_f = √970

v_f ≈ 31.1 m/s

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