(II)Roger sees water balloons fall past his window. Each balloon strikes the sidewalk 1 s after passing his window. His room is on the third floor, 15 m above the sidewalk. (a) How fast are the balloons traveling when they pass his window? (b) Assuming each balloon is released from rest, estimate the height from which it was released and the corresponding floor level. Each floor is 5 m high. Take g = 10 m/s².
Explanation
(a) Take downward as positive.
Using the equation of motion: s = ut + ½gt².
Here, s = 15 m, t = 1 s and g = 10 m/s².
15 = u(1) + ½(10)(1)².
15 = u + 5.
u = 10 m/s downward.
(b) The balloon was released from rest.
Using v² = u₀² + 2gΔh, with release velocity u₀ = 0:
The distance fallen before reaching the window is Δh = v²/(2g).
Δh = (10)² / (2 × 10) = 5 m.
Total release height above the sidewalk = 15 + 5 = 20 m.
Floor-height equivalent = 20 / 5 = 4.
This is an estimated floor-height level; the exact release floor depends on the building’s floor numbering and where the balloon was released.
✓
Conclusion
(a) The balloons pass Roger’s window at 10 m/s downward. (b) They were released from approximately 20 m above the sidewalk, equivalent to 4 floor-heights.
2
(II)A garden hose nozzle is pointed vertically upward from a height of 1.8 m above the ground. When the nozzle is quickly turned off, you hear the last water strike the ground after 2.5 s. Assuming air resistance is negligible, what was the speed of the water as it left the nozzle? Take g = 9.8 m/s².
Explanation
The last water leaving the nozzle continues moving upward before falling to the ground.
Let t be the time taken to reach maximum height.
Initial height h = 1.8 m; total flight time T = 2.5 s.
Initial speed: u = gt.
Maximum height above the ground: H = h + u²/(2g) = h + ½gt².
During the fall from maximum height, the water starts from rest.
The falling time is (T − t), so H = ½g(T − t)².
Equating the two expressions for H:
h + ½gt² = ½g(T − t)².
Simplifying: t = (T² − 2h/g)/(2T).
t = ((2.5)² − 2 × 1.8/9.8) / (2 × 2.5) = 1.18 s.
Therefore, u = gt = 9.8 × 1.18 = 11.53 m/s.
✓
Conclusion
The water leaves the nozzle at approximately 11.53 m/s upward.
3
(III)A baseball passes upward by a window with a vertical speed of 20 m/s. The ball was thrown by a person standing 18 m below the window on the street. (a) What was its initial speed? (b) What maximum altitude does it reach above the street? (c) How long before passing the window was it thrown? (d) When does it reach the street again, measured from launch? Take g = 10 m/s² and neglect air resistance.
Explanation
(a) Use the equation of motion v² = u² + 2as.
For upward motion, a = −g and displacement to the window is +18 m.
Therefore, v_window² = u² − 2gh_window.
u² = v_window² + 2gh_window.
u = √(20² + 2 × 10 × 18).
Initial speed u = 27.57 m/s.
(b) At maximum height, final velocity v = 0.
Using v² = u² − 2gH:
H = u²/(2g).
Maximum altitude above the street = 38 m.
(c) Using v = u − gt:
t = (u − v_window)/g.
Time before passing the window = 0.76 s.
(d) The ball returns to its launch height after the total flight time T = 2u/g.
T = 2 × 27.57 / 10.
Total time after launch = 5.51 s.
✓
Conclusion
(a) Initial speed = 27.57 m/s. (b) Maximum altitude = 38 m above the street. (c) The ball was thrown 0.76 s before passing the window. (d) It returns to the street 5.51 s after launch.
4
(III)A stone falling freely from rest takes 2 s to travel past a window 20 m tall. From what height above the top of the window was the stone dropped? Take g = 10 m/s² and neglect air resistance.
Explanation
Let u be the speed of the stone at the top of the window.
Using the equation of motion: s = ut + ½gt².
Substitute s = 20 m, t = 2 s and g = 10 m/s²:
20 = 2u + ½ × 10 × 2².
20 = 2u + 20.
Therefore, u = 0 m/s.
The stone was initially dropped from rest.
Using v² = u² + 2as, with initial velocity zero:
u_top² = 2gH.
H = u_top²/(2g).
H = (0)² / (2 × 10) = 0 m.
✓
Conclusion
The stone was dropped from a height of 0 m above the top of the window.
5
A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.4 s later. If the speed of sound is 340 m/s, how high is the cliff? Take g = 9.8 m/s² and neglect air resistance.
Explanation
The total time consists of two parts:
1. Time taken by the rock to fall.
2. Time taken by the sound to travel back to the top of the cliff.
Let the falling time be t and the cliff height be h.
For the falling rock, using s = ut + ½gt² with u = 0:
h = ½gt².
For the sound, time = distance/speed:
t_sound = h/v_sound.
Total time: T = t + h/v_sound.
Substitute h = ½gt²:
T = t + gt²/(2v_sound).
Rearranging gives: (g/(2v_sound))t² + t − T = 0.
Taking the positive root gives the physical falling time.
Falling time t ≈ 3.248 s.
Sound travel time ≈ 0.152 s.
Cliff height h = ½ × 9.8 × (3.248)² ≈ 51.69 m.
✓
Conclusion
The height of the sea cliff is approximately 51.69 m, accounting for both the rock’s falling time and the sound’s travel time.