Question

1. a 2.0 kg stone is thrown straight up into the air with a speed of 12 m/s from a cliff 55 m above the ocean. determine the speed of the stone when it hits the water. a. 210 m/s b. 35 m/s c. 33 m/s d. 32 m/s

Explanation:

Step1: Apply the conservation of mechanical energy

The initial mechanical energy \(E_1\) is the sum of kinetic energy \(K_1=\frac{1}{2}mv_1^{2}\) and potential energy \(U_1 = mgh_1\). The final mechanical energy \(E_2\) is the kinetic energy \(K_2=\frac{1}{2}mv_2^{2}\) (since \(h_2 = 0\) at the ocean level). According to \(E_1=E_2\), we have \(\frac{1}{2}mv_1^{2}+mgh_1=\frac{1}{2}mv_2^{2}\).

Step2: Simplify the equation

Divide the equation \(\frac{1}{2}mv_1^{2}+mgh_1=\frac{1}{2}mv_2^{2}\) by \(m\) (mass \(m\) cancels out). We get \(\frac{1}{2}v_1^{2}+gh_1=\frac{1}{2}v_2^{2}\).

Step3: Substitute the values

Given \(v_1 = 12\ m/s\), \(g = 9.8\ m/s^{2}\), \(h_1=55\ m\).

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Answer:

B. \(35\ m/s\)