Question

A large boulder is ejected vertically upward from a volcano with an initial speed of 39.5m/s . Air resistance may be ignored.

A. When is the displacement of the boulder from its initial position zero?
B. When is the velocity of the boulder zero?

Answer 1

The displacement of the boulder from its initial position will be zero when it reaches its maximum height and starts descending back down. The velocity of the boulder will also be zero at this point.

Step-by-step explanation:

Given that the boulder is ejected vertically upward from a volcano with an initial speed of 39.5 m/s and air resistance is ignored:

A. The displacement of the boulder from its initial position will be zero when it reaches its maximum height and starts descending back down.

B. The velocity of the boulder will be zero when it reaches its maximum height and starts descending back down.

In both cases, we can use the equations of motion to find the required values. For part A, we can use the equation:

vf^2 = vi^2 + 2aΔx

where vf is the final velocity (0 m/s), vi is the initial velocity (39.5 m/s), a is the acceleration (-9.8 m/s^2), and Δx is the displacement. Solving for Δx will give us the answer.

For part B, we can use the equation:

vf = vi + at

where vf is the final velocity (0 m/s), vi is the initial velocity (39.5 m/s), a is the acceleration (-9.8 m/s^2), and t is the time. Solving for t will give us the answer.

Answer 2

The boulder returns to the initial position at approximately
`8.05\; {\rm s}`.

The velocity of the boulder was momentarily zero at approximately
`4.03\; {\rm s}`.

Assumption:
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

Under the assumption that air resistance is negligible, the boulder would be accelerating at a constant
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}` in the vertical direction.

Since the rate of change in velocity is a constant value
`a`, the time
`t` required to achieve a velocity change of
`\Delta v` would be:

`\begin{aligned}t &= (\Delta v)/(a)\end{aligned}`,

Where
`\Delta v = (\text{current velocity}) - (\text{initial velocity})`.

As the boulder ascends, kinetic energy
`(\text{KE})` of the boulder is converted into gravitational potential energy
`(\text{GPE})`. When the boulder returns to the ground,
`(\text{GPE})\!` is converted back into
`(\text{KE})\!`.

Under the assumption that air resistance on the boulder is negligible, the sum of
`((\text{GPE}) + (\text{KE}))` should stay constant during the entire motion. Hence, when the boulder returns to the initial position,
`(\text{KE})\!` of the boulder should be the same as the initial value. While the boulder would be travelling at the same speed, the velocity of the boulder would be in the opposite direction:

• Initial velocity:
  `u = 39.5\; {\rm m\cdot s^(-1)}` (upward).
• Current velocity:
  `v = (-39.5)\; {\rm m\cdot s^(-1)}` (downward, and hence the negative sign.)

Divide the change in velocity by acceleration to find the duration of the motion:

`\begin{aligned}t &= (\Delta v)/(a) \\ &= \frac{(\text{current velocity})- (\text{initial velocity})}{a}\\ &= ((-39.5) - (39.5))/(9.81)\; {\rm s} \\ &\approx 8.05\; {\rm s}\end{aligned}`.

Apply the same equation to find the time required for the velocity of the boulder to be zero:

• Initial velocity:
  `u = 39.5\; {\rm m\cdot s^(-1)}`.
• Current velocity:
  `v = 0\; {\rm m\cdot s^(-1)}`.

Hence:

`\begin{aligned}t &= (\Delta v)/(a) \\ &= \frac{(\text{current velocity})- (\text{initial velocity})}{a}\\ &= (0 - 39.5)/(9.81)\; {\rm s} \\ &\approx 4.03\; {\rm s}\end{aligned}`.