
PUZZLE SOLUTIONSfor 
1. Let the officer on the horse be "O"
2. Let the rider at the rear of the column be “R"
3. The velocity of the officer to the ground (v_{OG}) = velocity of officer to rear (v_{OR}) plus velocity of rear with respect to ground (v_{RG}). That is, v_{OG} = v_{OR} + v_{RG}
4. Let time for O to reach head of column be t_{1}. Hence t_{1} = s_{1}/v_{OG}
5. In this time R has moved a distance s_{2}. Hence t_{1} = s_{2}/v_{RG}
6. Both are equal to t_{1}, hence: s_{1}/v_{OG} = s_{2}/v_{RG}
7. But s_{1}= 3 + s_{2} thus (3 + s_{2})/v_{OG} = s_{2}/v_{RG}
8. Rearranging:(3 + s_{2})/s_{2} = v_{OG}/v_{RG}
9 Let the distance O travels from the head of the column to the rear be s_{3}
10. Let the time for O to travel from head of column back to rear = t_{2} (= s_{3}/v_{OG})
11. In this time R has moved a distance s_{4} in a time t_{2} (= s_{4}/v_{RG})
12. But s_{4} = 4  s_{2}
13. Substituting into equation in line 11: t_{2} = (4  s_{2})/v_{RG}
14. But s_{4} = 3  s_{3} = 4  s_{2}. Hence s_{3} = s_{2} 1.
15. Substituting into line 10: t_{2} = s_{3}/v_{OG} = (s_{2}  1)/v_{OG}
16. Hence t_{2} = (s_{2}  1)/v_{OG} = (4  s_{2})/v_{RG}
17. Rearranging: v_{OG}/v_{RG } = (s_{2 } 1)/(4  s_{2}), but this equals (3 + s_{2})/s_{2} from line 8.
18. Hence (s_{2}  1)/(4  s_{2}) = (3 + s_{2})/s_{2}
19. Rearranging: (s_{2}  1)s_{2} = (3 + s_{2})(4  s_{2})
20. s_{2}^{2}  s_{2} = 12  3s_{2} + 4s_{2}  s_{2}^{2}
21. 2s_{2}^{2}  2s_{2} 12 = 0
22. Simplify: s_{2}^{2}  s_{2}  6 = 0
23. Factorise: (s_{2}  3)(s_{2} + 2) = 0. Hence s_{2} = 3 (ignore 2)
24. If s_{2} = 3, then s_{1} = 6, s_{3} = 2, s_{4} = 1
THUS  the officer travels s_{1} + s_{3} = 6 + 2 = 8 km
By the way, the velocity of the officer and column can be found:
25. Substituting into equation 7: v_{RG} = v_{OG}/2
26. But v_{OG} = v_{OR} + v_{RG}, hence 2v_{RG} = v_{OR} + v_{RG}, thus v_{RG} = v_{OR}
27. In the total time of t_{1} + t_{2}, R has traveled 4 km, hence: v_{RG} = 4/(t_{1}+t_{2})
28. Substituting for t_{1} (equation 4) and t_{2} (equation 15): (using actual values of s_{1} and s_{3}):
v_{RG} = 4 /((6/v_{OG}) + 2/v_{OG})) = 8/v_{OG}
29. Hence v_{RG} = 8/v_{OG} = v_{OG}/2 (from equation 25)
30. v_{OG}^{2} = 16, hence v_{OG} = 4, and v_{RG} = 2.
The officer travels at 4 km/h and the column travels at 2 km/h.
The times are t_{1} = 1.5 hr, t_{2} = 0.5 hr. Total time 2 hours.
Alternative solution provided by Steve Shellshear (you’ve really got to get out more Steve).
Initial velocity = v
Angle of projection = q
Initial vertical velocity component u_{v}= vsinq
Initial horizontal velocity component v_{H}= vcosq
Travel 5 m (s_{H}) horizontally in time t:
Rearranging v_{H} = s_{H}/t gives s_{H} = v_{H} t = vcosq
Substituting: 5 = vcosq t
Rearranging: t = 5/ vcosq (EQUATION 1)
Travel vertically half maximum height (s_{v} = h_{max}/2) in time t:
Using s_{v} = u_{v}t + ½at^{2}
h_{max}/2 = vsinq t + ^{}5t^{2} (EQN 2)
Substitute Eq 1 into Eq 2:
h_{max}/2 = vsinq 5/ vcosq + ^{}5(5/ vcosq)^{2}
Multiply both sides by 2:
h_{max}/2 = 10sinq/cosq + ^{}250/(vcosq)^{2} (EQN 3)
Determine h_{max }using v_{v}^{2} = u_{v}^{2} + 2as
0 = (vsinq)^{2} – 20 h_{max}
Rearranging:
20h_{max} = (vsinq)^{2}
Divide both sides by 20:
h_{max} = (vsinq)^{2}/20 (EQN 4)
Substitute EQN 4 into EQN 3:
(vsinq)^{2}/20 = 10sinq/cosq + ^{}250/(vcosq)^{2}
Multiply both sides by 20 (vcosq)^{2}
(vsinq)^{2}(vcosq)^{2} = 200sinq (vcosq)^{2}/cosq – 250 x 20
v^{4} (sinqcosq)^{2} = 200sinqcosq v^{2}  5000
Prepare for identity conversion of 2 sinqcosq = sin2q
v^{4} (2sinqcosq)^{2}/4 = 100 x 2sinqcosq v^{2}  5000
Convert and multiply both sides by 4:
v^{4} (sin2q)^{2} = 400 v^{2} sin 2q  20000
Subtract 400v^{2}sin2q from, and add 20000 to both sides:
v^{4} (sin2q)^{2} – 400v^{2}sin2q + 20000 = 0 (Equation 5)
Calculate the range using time t
Vertically, using s_{v} = u_{v}t + ½at^{2}
0 = vsinq t – 5t^{2}
Add 5t^{2} and subtract vsinqt from both sides:
5t^{2}  vsinqt = 0
Divide both sides by 5:
t^{2}  vsinqt/5 = 0
Factorise:
T(tvsinq/5) = 0
Therefore, t = o t = vsinq/5
We are interested in t = vsinq/5 (EQN 6)
Horizontal range, R:
R = vcosqt (EQN 7)
R = vcosq vsinq/5
Convert to sin 2q
R = v^{2} (2sinq cosq)/10
R = v^{2} sin2q/10 (EQN 8)
Substitute EQN 8 into EQN 5
100R^{2} – 4000R + 20000 = 0
Divide both sides by 100
R^{2} – 40R + 200 = 0
Using quadratic formula:
R = (40 ± Ö(40^{2} – 4 x 1 x 200)/(2 x 1)
R = (40
± 28.284)/2
R =
34.14m or 5.86m
Thanks Steve.