Determine the time it takes the rock to follow this path. What is the speed and angle below the positive horizontal which the rock impacts the side of the volcano?

Question

Kilauea in Hawaii is the world's most continuously active volcano. Very active volcanoes characterisitcally eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 29m/s29m/s and at an angle of 37∘37∘ above the horizontal, as shown in the figure below. The rock strikes the ground at an altitude of 20m20m lower than its starting point. Include units as appropriate below.

(a) Determine the time it takes the rock to follow this path.

𝑡t = 

(b) What is the speed and angle below the positive horizontal which the rock impacts the side of the volcano?

𝑣𝑓 =  ,    𝜃 = 

Answer 1

Let

g = 9.81 m/s² be gravitation acceleration,

v(t) be the (vector) velocity of the rock after some time t.

So v(0) = v₀.

Let t₁ be the time it takes the rock to hit the ground.

The assignment is to compute t₁ and v(t₁).

v₀ can be decomposed into a vector sum of

its horizontal component of magnitude v₀cos(θ₀) and

its vertical component of magnitude v₀sin(θ₀).

While the horizontal component remains constant, the vertical component

is constantly reduced by gravity.

So after some time t the vertical component of v(t) is

v₀sin(θ₀) - gt (1)

The horizontal component of position (with respect to the starting position) is

v₀cos(θ₀)t

because horizontally the rock moves at constant velocity.

The vertical position of the rock (with respect to the starting position) is

v₀sin(θ₀)t - gt²/2

(If you do not know where this comes from let me know.)

We are given that at the sought time t₁ the vertical position is -20, so

v₀sin(θ₀)t₁ - gt₁²/2 = -20

Substituting actual numbers we get the quadratic equation

4.9t₁² - 17.4t₁ - 20 = 0

t₁ = (17.4 ± √(17.4² + 196.2))/9.81

This gives us two solutions

t₁ = -0.5, and t₁ = 4

The solution t₁ = -0.5 does not correspond to our physical constraints,

while t₁ = 4s is the answer to question (a)

We can plug that into (1) to get the vertical component of the velocity v(t₁) as

29×sin(37°) - 9.81×4 = -21.8 m/s

Note that it is negative, because it is oriented downward.

The horizontal component of v(t₁) is the constant

v₀cos(θ₀) = 29×cos(37°) = 23.2 m

The tangent of θ is the ration between the two components

tan(θ) = -21.8/23.2 = -0.94

Therefore θ = -43.2°

(The angle is negative because it points below the horizontal.)

The magnitude of the velocity (i.e., speed) is the Euclidean sum of the two components

|v(t₁)| = √(21.8² + 23.2²) = 31.8 m/s