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Difficulty: Very hardKinematics and Linear Motion

A body is released from rest from the top of a cliff of height hh. If it covers a distance equal to 716h\frac{7}{16}h in the final second of its motion before hitting the ground, what is the total height hh of the cliff? (Take acceleration due to gravity g=10 m s−2g = 10\text{ m s}^{-2})

  1. A
    45 m45\text{ m}
  2. 80 m80\text{ m}Answer
  3. C
    100 m100\text{ m}
  4. D
    160 m160\text{ m}

Answer

80 m80\text{ m}
Using the equation of motion under constant acceleration, the total distance fallen from rest in time tt is h=12gt2h = \frac{1}{2}gt^2. The distance fallen up to time (t−1)(t-1) is ht−1=12g(t−1)2h_{t-1} = \frac{1}{2}g(t-1)^2. The distance traveled during the final second is Δh=h−ht−1=12g(2t−1)\Delta h = h - h_{t-1} = \frac{1}{2}g(2t-1). Equating this to 716h\frac{7}{16}h yields 12g(2t−1)=716(12gt2)\frac{1}{2}g(2t-1) = \frac{7}{16}\left(\frac{1}{2}gt^2\right), which simplifies to 7t2−32t+16=07t^2 - 32t + 16 = 0. Factoring gives t=4 st = 4\text{ s} (rejecting t=47 st = \frac{4}{7}\text{ s} since time must exceed 1 s1\text{ s}). Substituting t=4 st = 4\text{ s} into h=12(10)(4)2h = \frac{1}{2}(10)(4)^2 gives 80 m80\text{ m}.

Step-by-Step Solution

1
Express total height hh in terms of total fall time tt.
h=12gt2=5t2h = \frac{1}{2} g t^2 = 5t^2
Since the body starts from rest (u=0 m s−1u = 0\text{ m s}^{-1}), displacement under uniform acceleration g=10 m s−2g = 10\text{ m s}^{-2} is given by h=12gt2h = \frac{1}{2}gt^2.
2
Express the height fallen in the first (t−1)(t - 1) seconds.
h′=12g(t−1)2=5(t−1)2h' = \frac{1}{2} g (t - 1)^2 = 5(t - 1)^2
The distance covered up to one second before impact is the total distance fallen minus the distance covered in the final second.
3
Calculate the distance fallen in the final second and set up the equation.
Δh=h−h′=5t2−5(t−1)2=5(2t−1)\Delta h = h - h' = 5t^2 - 5(t - 1)^2 = 5(2t - 1). Given Δh=716h\Delta h = \frac{7}{16}h, we have 5(2t−1)=716(5t2)5(2t - 1) = \frac{7}{16}(5t^2).
The distance fallen during the last second is the difference between total height and height fallen up to (t−1)(t-1) seconds.
4
Solve the quadratic equation for tt.
7t2−32t+16=0  ⟹  (7t−4)(t−4)=0  ⟹  t=4 s7t^2 - 32t + 16 = 0 \implies (7t - 4)(t - 4) = 0 \implies t = 4\text{ s} (since t>1 st > 1\text{ s}).
Simplifying 2t−1=716t22t - 1 = \frac{7}{16}t^2 gives 7t2−32t+16=07t^2 - 32t + 16 = 0. The root t=4/7 st = 4/7\text{ s} is discarded as tt must be greater than 1 s1\text{ s}.
5
Substitute t=4 st = 4\text{ s} back into the total height formula.
h=5(4)2=80 mh = 5(4)^2 = 80\text{ m}.
Calculating total height using h=5t2h = 5t^2 for t=4 st = 4\text{ s} gives 80 m80\text{ m}.

Key Concept

Free Fall under Gravity and Motion in the nn-th Second
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Topics

Kinematics and Linear Motion

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