To solve this problem, we can use the equations of motion for an object under constant acceleration. In this case, the acceleration is due to gravity, and its value is approximately 9.8 m/s² (assuming no air resistance).
Let's analyze the motion of the stone in two parts: the upward motion and the downward motion.
- Upward motion: The initial velocity of the stone is +12.0 m/s (upward), and the final velocity at the highest point is 0 m/s (at the instant it changes direction). We can use the following equation to find the time taken to reach the highest point:
v = u + at
0 = 12.0 - 9.8t
Solving for t, we get: t = 12.0 / 9.8 ≈ 1.22 s
Using this time, we can find the maximum height reached by the stone using the equation:
s = ut + (1/2)at²
s = 12.0 * 1.22 - (1/2) * 9.8 * (1.22)²
s ≈ 14.7 m
- Downward motion: Now, the stone starts from rest at a height of 14.7 m and falls freely under gravity. We can use the equation of motion for vertical displacement to find the time taken to fall from this height:
s = ut + (1/2)at²
-70 = 0 * t + (1/2) * 9.8 * t²
35t² = 70
t² = 70 / 35
t² = 2
t ≈ √2 ≈ 1.41 s
Now that we have the total time of flight, we can find the final velocity when the stone hits the ground using the equation:
v = u + at
v = 0 + 9.8 * 1.41
v ≈ 13.78 m/s
Therefore, the magnitude of the speed just before hitting the ground at the base of the cliff is approximately 13.78 m/s.