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
Answer:
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
