Question

The first active volcano observed outside the earth was discovered in 1979 on io, one of the moons of jupiter. suppose a volcano on another moon is observed to be ejecting material to a height of about 1.52 ✕ 105 m. given that the acceleration of gravity on this moon is 1.91 m/s2 find the initial velocity of the ejected material.

Answer 1

The initial velocity of the ejected material from the volcano can be found by using the kinematic equation which, when the values are substituted, results in an initial velocity of approximately 761.6 m/s.

Step-by-step explanation:

The question involves finding the initial velocity of material ejected from a volcano on a moon similar to Io, given the height reached by the ejected material and the moon's acceleration due to gravity. To calculate the initial velocity of the ejected material, we can use the kinematic equation that relates velocity, acceleration, and displacement under constant acceleration, which is appropriate since we are ignoring air resistance and other forces.

Using the equation v2 = u2 + 2as, where v is the final velocity (which is zero at the highest point of projection), u is the initial velocity, a is the acceleration due to gravity (negative in this case since it's acting opposite to the motion), and s is the maximum height reached. Rearranging for u, we get u = √(-2as).

Plugging in the values: s = 1.52 x 105 m, and a = -1.91 m/s2, we get:
u = √(-2 × -1.91 m/s2 × 1.52 x 105 m)
= √(5.7984 x 105 m2/s2).

Hence, the initial velocity, u, is the square root of 5.7984 x 105, which is approximately 761.6 m/s.

Answer 2

Gravitation due to the planet's mass is always against the ejected particle velocity. This will result in deceleration of the ejected particle and eventually bring the velocity to zero at the topmost point of the trajectory before pulling it back to the ground.

By the kinematic equation

`v^2 - u^2 = 2.a.s`

where s is the distance travelled = 1.52 x 10^5 m

a is the acceleration due to gravity = -1.91 m/s^2

v and u are initial and final velocities, so v = 0.

Therefore

`u = √(s.a.s)`

u = 761.99 m/s