Question

A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignore and using conservation of mechanical energy, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.

Answer 1

a) The initial speed of the rock is approximately 14.607 meters per second.

b) The greatest height of the rock from the base of the cliff is 42.878 meters.

Step-by-step explanation:

a) The rock experiments a free-fall motion, that is a vertical uniform accelerated motion due to gravity, in which air friction and effects of Earth's rotation. By Principle of Energy Conservation we have the following model:

`U_(g,1)+K_(1) = U_(g,2)+K_(2)` (Eq. 1)

Where:

`U_(g,1)`,
`U_(g,2)` - Initial and final gravitational potential energies, measured in joules.

`K_(1)`,
`K_(2)` - Initial and final translational kinetic energies, measured in joules.

By definitions of gravitational potential and translational kinetic energies we expand and simplify the equation above:

`m\cdot g\cdot (y_(1)-y_(2))= (1)/(2)\cdot m\cdot (v_(2)^(2)-v_(1)^(2))`

`g\cdot (y_(1)-y_(2)) = (1)/(2)\cdot (v_(2)^(2)-v_(1)^(2))` (Eq. 2)

Where:

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

`y_(1)`,
`y_(2)` - Initial and final height, measured in meters.

`v_(1)`,
`v_(2)` - Initial and final speed of the rock, measured in meters per second.

If we know that
`g = 9.807\,(m)/(s^(2))`,
`y_(1) = 32\,m`,
`y_(2) = 0\,m` and
`v_(2) = 29\,(m)/(s)`, then the equation is:

`\left(9.807\,(m)/(s^(2)) \right)\cdot (32\,m-0\,m) = (1)/(2)\cdot \left[\left(29\,(m)/(s) \right)^(2)-v_(1)^(2)\right]`

`313.824 = 420.5-0.5\cdot v_(1)^(2)`

`0.5\cdot v_(1)^(2) = 106.676`

`v_(1) \approx 14.607\,(m)/(s)`

The initial speed of the rock is approximately 14.607 meters per second.

b) We use (Eq. 1) once again and if we know that
`g =9.807\,(m)/(s^(2))`,
`y_(1) = 32\,m`,
`v_(1) \approx 14.607\,(m)/(s)` and
`v_(2) = 0\,(m)/(s)`, then the equation is:

`\left(9.807\,(m)/(s^(2)) \right)\cdot (32\,m-y_(2)) = (1)/(2)\cdot \left[\left(0\,(m)/(s) \right)^(2)-\left(14.607\,(m)/(s) \right)^(2)\right]`

`313.824-9.807\cdot y_(2) = -106.682`

`9.807\cdot y_(2) = 420.506`

`y_(2) = 42.878\,m`

The greatest height of the rock from the base of the cliff is 42.878 meters.