Question 1 · Hard
A wide open tank stands on level ground with its water surface kept 5 m above the ground. Water leaves through a small hole in the side wall, 1.8 m above the ground. Taking g = 10 m/s², how far from the base of the wall does the jet strike the ground?
- A4.8 m
- B2.4 m
- C3.6 m
- D6.0 m
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Answer: A · 4.8 m
The hole is 3.2 m below the surface, so v = √(2 × 10 × 3.2) = 8 m/s. Fall time = √(2 × 1.8/10) = 0.6 s, so the range = 8 × 0.6 = 4.8 m. Check: range = 2√(3.2 × 1.8) = 2 × 2.4 = 4.8 m.
- B: This uses √(h₁h₂) without the factor 2.
- C: This takes the head as the 1.8 m height of the hole above the ground.
- D: This takes the head as the full 5 m.
Remember this
Range = 2√(h₁h₂), where h₁ is the depth of the hole and h₂ its height above the ground.