A climber ascends a vertical rock face at 5 m/min for the first 3 minutes, then at 3 m/min thereafter. From the bottom of the cliff, the angle of elevation to the climber is measured every minute. If at t=0 the angle is 30° and at t=7 min it is 60°, determine the cliff’s height.
Show answer & explanation
Correct answer: A
Explanation
Let the height of the cliff = H meters, and the horizontal distance of the observer from the base of the cliff = x meters.
At t = 0, the climber is at a height of h meters from the base of the cliff.
Climb in the first 3 minutes = 3 × 5 = 15 meters
Climb in the next 4 minutes = 4 × 3 = 12 meters
Therefore, total climb in 7 minutes = 27 meters
tan30° = h/x
h = x/√3
tan60° = (h + 27)/x
h + 27 = √3x
Substituting h = x/√3:
x/√3 + 27 = √3x
x + 27√3 = 3x
2x = 27√3
x = 27√3/2
h = x/√3 = 13.5 meters
Height of the climber at 7 minutes = 13.5 + 27 = 40.5 meters