19. a rock is thrown off a cliff, 123 m high, at an angle of 32 degrees at a speed of 20 m/s. at what speed…

Explanation:

Step1: Apply the conservation of mechanical energy

The initial mechanical energy (E_{i}) is the sum of kinetic energy (K_{i}=\frac{1}{2}mv_{i}^{2}) and potential energy (U_{i} = mgh). The final mechanical energy (E_{f}) is just kinetic energy (K_{f}=\frac{1}{2}mv_{f}^{2}) (since (h = 0) at the ground level). According to the conservation of mechanical energy (E_{i}=E_{f}), so (\frac{1}{2}mv_{i}^{2}+mgh=\frac{1}{2}mv_{f}^{2}).

Step2: Solve for (v_{f})

Cancel out the mass (m) from each term of the equation (\frac{1}{2}v_{i}^{2}+gh=\frac{1}{2}v_{f}^{2}). We know that (v_{i} = 20\space m/s), (h=123\space m), and (g = 9.8\space m/s^{2}). First, calculate (\frac{1}{2}v_{i}^{2}=\frac{1}{2}\times(20)^{2}= 200\space m^{2}/s^{2}) and (gh=9.8\times123 = 1205.4\space m^{2}/s^{2}). Then (\frac{1}{2}v_{f}^{2}=200 + 1205.4=1405.4\space m^{2}/s^{2}). Multiply both sides by (2) to get (v_{f}^{2}=2\times1405.4 = 2810.8\space m^{2}/s^{2}). Take the square - root: (v_{f}=\sqrt{2810.8}\approx53\space m/s).

Answer:

The speed at which the rock hits the ground is approximately (53\space m/s).