Mt. Asama, Japan, is an active volcano complex. In 2009 , an eruption threw solid volcanic rocks that landed far from the crater. Suppose hat one such rock was launched at an angle of θ=3.5 degrees above horizontal, and landed a horizontal distance d=680 m from the crater, and a vertical listance h=470 m below the crater. Write and expression for ν
0

, the initial speed of the rock in terms of g,θ,d, and h, What is the initial speed of the rock in m/s?

Answers

Answer 1

The initial speed of the rock is approximately 190.66 m/s.

To find the initial speed of the rock (v0), we can use the kinematic equations for projectile motion. Given the angle (θ), horizontal distance (d), and vertical distance (h), we can express v0 in terms of g, θ, d, and h.

The horizontal distance traveled by the rock can be expressed as:

d = v0 * cos(θ) * t

The vertical distance traveled can be expressed as:

h = v0 * sin(θ) * t - (1/2) * g * t²

We can solve the first equation for t:

t = d / (v0 * cos(θ))

Now, substitute this expression for t in the second equation:

h = v0 * sin(θ) * (d / (v0 * cos(θ))) - (1/2) * g * (d / (v0 * cos(θ)))²

Simplifying:

h = (v0 * sin(θ) * d) / (v0 * cos(θ)) - (1/2) * g * (d² / (v0² * cos²(θ)))

Simplify further:

h = (sin(θ) * d) / cos(θ) - (1/2) * g * (d² / (v0² * cos²(θ)))

Multiply through by cos(θ):

h * cos(θ) = sin(θ) * d - (1/2) * g * (d² / v0²)

Rearranging the equation:

(h * cos(θ)) - (sin(θ) * d) = (1/2) * g * (d² / v0²)

Now, solve for v0:

v0² = (2 * g * (d² / ((h * cos(θ)) - (sin(θ) * d))))

Taking the square root to isolate v0:

v0 = sqrt(2 * g * (d² / ((h * cos(θ)) - (sin(θ) * d))))

Given the values:

θ = 3.5 degrees

d = 680 m

h = 470 m

g = 9.8 m/s²

Convert θ to radians:

θ_rad = 3.5 * (π/180) ≈ 0.0611 rad

Substitute the values:

v0 = sqrt(2 * 9.8 * (680² / ((470 * cos(0.0611)) - (sin(0.0611) * 680))))

Calculating the above expression gives:

v0 ≈ 190.66 m/s

Therefore, the initial speed of the rock is approximately 190.66 m/s.

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Related Questions

In which ways may a fault be recognized in a landscape?
-fault scarp
-offset beds
-fault breccia

Answers

Faults are fractures that extend into the earth's crust and, in some instances, the mantle. Faulting has three principal types, each with distinct identifying characteristics and consequences.

Fault scarp, offset beds, and fault breccia are three ways in which a fault may be recognized in a landscape.What are fault scarps?A fault scarp is a linear vertical or near-vertical wall-like feature that develops along the earth's surface due to movement on the underlying fault. The surface trace of a fault is represented by a fault scarp, which is the outcrop or topographic expression of the fault plane in the surface.

The higher and steeper the scarp, the greater the relative motion on the fault.What are offset beds?An offset bed is a sedimentary layer or stratum that has been broken and displaced by faulting. The outcrop pattern of these layers is interrupted and does not match from one side of the fault to the other. In order to create a fault, one part of the ground must be shifted with respect to the other. This offset is noticeable on the exposed beds that appear on the fault surface.

The relative displacement of the rock layers that are divided by the fault is referred to as fault offset.What is fault breccia?Fault breccia is a type of rock that is created when rocks on either side of a fault are pulverized and mixed together as a result of the high stress associated with faulting. When a rock is fractured, it can be broken into tiny fragments known as breccia.

Fault breccia is a type of rock that is formed when the high stress associated with faulting crushes and grinds rock on either side of the fault until it becomes pulverized and is mixed together.

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