A stone is dropped into a river from a bridge 43.9 m above the water. Another stone is thrown vertically downward 1 s later. Both stones strike the water at the same time. (a) What is the initial speed of the second stone? (b) Describe the velocity-versus-time graph for both stones, taking t = 0 as the instant the first stone is released. Take g = 9.8 m/s² and neglect air resistance.
Explanation
(a) First stone: it is dropped from rest.
Using the equation of motion h = ½gt₁²:
t₁ = √(2h/g) = √(2 × 43.9/9.8) = 2.993 s.
The second stone is thrown 1.00 s later.
Its falling time is t₂ = t₁ − 1 = 1.993 s.
For the second stone, use the equation of motion h = ut + ½gt².
Rearranging: u = (h − ½gt²)/t.
u₂ = (43.9 − ½ × 9.8 × (1.993)²)/1.993.
Initial speed of the second stone = 12.26 m/s downward.
(b) Take downward as positive.
Stone 1: v₁ = gt, from t = 0 until t = t₁.
Its velocity graph is a straight line starting at (0, 0) with slope +9.8 m/s².
Stone 2: v₂ = 0 until t = 1.00 s, then v₂ = u₂ + g(t − 1.00).
Its graph is horizontal at zero until 1.00 s, then a straight line with the same slope as stone 1.
Both stones reach the water at t = 2.993 s.
✓
Conclusion
(a) The second stone must be thrown downward at approximately 12.26 m/s. (b) The first stone’s velocity increases linearly from zero; the second stone’s velocity begins increasing 1.00 s later. Both have the same constant acceleration of 9.8 m/s² downward.
2
A ball of moist clay falls 15 m to the ground. It remains in contact with the ground for 20.0 ms before stopping. (a) What is the magnitude of its average acceleration while in contact with the ground? (b) Is the average acceleration upward or downward? Take g = 9.8 m/s² and treat the ball as a particle.
Explanation
Step 1: Find the speed immediately before impact.
The ball falls from rest, so u = 0.
Using the equation of motion: v² = u² + 2gh.
v = √(2 × 9.8 × 15) = 17.15 m/s downward.
Step 2: Find the average acceleration during contact.
Convert the contact time into seconds:
Δt = 20.0 ms = 0.0200 s.
Take upward as positive.
Initial velocity during contact = −17.15 m/s.
Final velocity = 0 m/s.
Using average acceleration: a_avg = (v_f − v_i)/Δt.
a_avg = (0 − (−17.15)) / 0.0200.
a_avg = 857.32 m/s².
Step 3: Determine the direction.
The ball’s velocity changes from downward to zero.
Therefore, its average acceleration is upward, opposite to its velocity just before impact.
✓
Conclusion
(a) The magnitude of the average acceleration is 857.32 m/s². (b) The average acceleration is upward.