Question

9) A rock is projected upwards from the surface of the moon at time t = 0.0 s with a velocity of 30 m/s (the acceleration due to gravity at the surface of the moon is -1.62 m/s2. The rock rises, then falls and strikes the ground. Consider all quantities as positive in the upward direction. What is the velocity during accent at 180 m above the surface? At time t = 10 s, what is the height of the rock from the ground?​

Answer 1

a) The velocity of the rock during ascent at 180 meters above the surface is approximately 17.799 meters per second.

b) The height of the rock from the ground is 219 meters from the ground.

Step-by-step explanation:

a) The rock experiments a free fall, in which this element is projected upwards, decelerated until rest is reached and then it falls down. Please notice that free fall is a case of uniformly accelerated motion, in which object is accelerated by gravity. The velocity of the rock is described by the following kinematic equation:

`v = \sqrt{v_(o)^(2)+2\cdot a\cdot \Delta s}` (1)

Where:

`v_(o)` - Initial velocity, measured in meters per second.

`v` - Final velocity, measured in meters per second.

`a` - Gravitational acceleration, measured in meters per square second.

`\Delta s` - Travelled distance, measured in meters.

If we know that
`v_(o) = 30\,(m)/(s)`,
`a = -1.62\,(m)/(s^(2))` and
`\Delta s = 180\,m`, then the velocity during ascent at 180 meters is:

`v =\sqrt{\left(30\,(m)/(s) \right)^(2)+2\cdot \left(-1.62\,(m)/(s^(2))\right)\cdot (180\,m) }`

`v \approx 17.799\,(m)/(s)`

The velocity of the rock during ascent at 180 meters above the surface is approximately 17.799 meters per second.

b) The height of the rock is calculated from this kinematic formula:

`s = s_(o)+v_(o)\cdot t +(1)/(2)\cdot a \cdot t^(2)` (2)

Where:

`s_(o)` - Initial position of the rock, measured in meters.

`s` - Final position of the rock, measured in meters.

`v_(o)` - Initial velocity of the rock, measured in meters per second.

`t` - Time, measured in seconds.

`a` - Acceleration, measured in meters per square second.

If we know that
`s_(o) = 0\,m`,
`v_(o) = 30\,(m)/(s)`,
`t = 10\,s` and
`a = -1.62\,(m)/(s^(2))`, then the height of the rock from the ground is:

`s = 0\,m + \left(30\,(m)/(s) \right)\cdot (10\,s)+(1)/(2)\cdot \left(-1.62\,(m)/(s^(2)) \right)\cdot (10\,s)^(2)`

`s = 219\,m`

The height of the rock from the ground is 219 meters from the ground.