Question

a rock is kicked horizontally at a speed of 15.0 m/s from the edge of a cliff. the rock strikes the ground 60.0 m from the foot of the cliff. neglecting air resistance, determine the height of the cliff in meters.

Answer 1

Approximately
`78.5\; {\rm m}` (assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

Under the assumption that air resistance is negligible, while the rock was in the air:

• Horizontal velocity would be constant:
  `v_(x) = 15.0\; {\rm m\cdot s^(-1)}`, same as the initial value.
• Vertical velocity would be changing with a constant vertical acceleration of
  `a_(y) = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`.

Let
`u_(y)` denote the initial vertical velocity of this rock. Since the rock was launched horizontally,
`u_(y) = 0\; {\rm m\cdot s^(-1)}`. Given that the horizontal distance (range) travelled is
`s = 60.0\; {\rm m}`, the height of this cliff can be found in the following steps:

• Find the duration of the flight from the motion in the horizontal direction.
• Apply the SUVAT equation
  `x_(y) = (1/2)\, a_(y)\, t^(2) + u_(y)\, t` to find the change in the position of the rock in the vertical direction. The height of the cliff would be equal to the absolute value of this vertical displacement.

Since horizontal velocity is constantly
`v_(x) = 15.0\; {\rm m\cdot s^(-1)}`, duration of the flight can be found by dividing the horizontal displacement by this horizontal velocity:

`\displaystyle t = (s)/(v_(x))`.

Since the acceleration in the vertical direction is constant, changes in the vertical position can be found using the SUVAT equation
`x_(y) = (1/2)\, a_(y)\, t^(2) + u_(y)\, t`:

`\begin{aligned}x_(y) &= (1)/(2)\, a_(y)\, t^(2) + u_(y)\, t \\ &= (1)/(2)\, a_(y)\, \left((s)/(v_(x))\right) ^(2) + u_(y)\, \left((s)/(v_(x))\right) \\ &= (1)/(2)\, (-9.81)\, \left((60.0)/(15.0)\right) + (0)\, \left((60.0)/(15.0)\right) \\ &\approx (-78.5)\; {\rm m}\end{aligned}`.

(The value of vertical displacement is negative because the new position of the rock is below where it was launched.)

Hence, the height of the cliff would be approximately
`78.5\; {\rm m}`.