EXAMPLE 8-3 Falling rock. If the original height of the rock in Fig. 8-7 is y1 = h = 3.0 m, calculate the rock's speed when it has fallen to 1.0 m above the ground.
APPROACH We apply the principle of conservation of mechanical energy, Eq. 8-12, with only gravity acting on the rock. We choose the ground as our reference level (y = 0).
SOLUTION At the moment of release (point 1) the rock's position is y1 = 3.0 m and it is at rest: v1 = 0. We want to find v2 when the rock is at position y2 = 1.0 m. Equation 8-12 gives
1/2mv1^2 + mgy1 = 1/2mv2^2 + mgy2.
The m's cancel out; setting v1 = 0 and solving for v2 we find
v2 = sqrt(2g(y1 - y2)) = sqrt(2(9.8 m/s^2)[(3.0 m) - (1.0 m)]) = 6.3 m/s.
The rock's speed 1.0 m above the ground is 6.3 m/s downward.
NOTE The velocity of the rock is independent of the rock's mass.
EXERCISE C In Example 8-3, what is the rock's speed just before it hits the ground? (a) 6.5 m/s; (b) 7.0 m/s; (c) 7.7 m/s; (d) 8.3 m/s; (e) 9.8 m/s.
Equation 8-12 can be applied to any object moving without friction under the action of gravity. For example, Fig. 8-8 shows a roller-coaster car starting from rest at the top of a hill, and coasting without friction to the bottom and up the hill on the other side.