Question

The trajectory of a rock ejected from the Kilauea volcano, with a velocity of magnitude 6.4 m/s and at angle 2.2 degrees above the horizontal. The rock strikes the side of the volcano at an altitude of 1.4m lower than its starting point. Calculate the magnitude of the rock's velocity at impact? Use g = 9.8 m/s2.

Answer 1

Answer:

v = 8.03 m / s

Explanation:

This is a missile throwing exercise.

          y = y₀ + $v_{oy}$  t - ½ g t²

     

indicates that y = -1.2 m and the initial velocity is

         v_{oy} = v₀ sin θ

         v_{oy} = 6.4 sin 2.2

         v_{oy} = 0.2457 m / s

we substitute in the equation

         -1.2 = 0.2457 t - ½ 9.8 t²

          4.9 t² - 0.2457 t - 1.2 = 0

          t² - 0.05014 t -0.2449 = 0

we solve the quadratic equation

          t = [0.05014 ±√ (0.05014² + 4 0.2449)] / 2

          t = [0.05014 ± 0.9910] / 2

          t₁ = 0.5206 s

          t₂ = -0.47 s

time must be a positive magnitude therefore the correct answer is

          t = 0.5206 s

with this time we can calculate the vertical speed when the rock hits the ground

         v_{y} = v_{oy} - gt

         v_{y} = 0.2457 - 9.8 0.5206

         v_{y} = -4.856 m / s

the negative sign indicates that the speed is down

horizontal velocity is constant, due to no acceleration

         vₓ = v₀ₓ = v₀ cos 2,2

         v₀ₓ = 6.4 cos 2.2

         v₀ₓ = 6.395 m / s

therefore let's use Pythagoras' theorem to find the velocity

         v = √ (vₓ² + $v_{y}^{2}$)

         v = √ (6,395² + 4,856²)

         v = 8.03 m / s

the direction can be found with trigonometry

        tan θ = v_{y} / vₓ

        θ = tan⁻¹ (-4,856 / 6,395)

        θ = - 37

the negative sign indicates that it is half clockwise from the x axis