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An object falls freely from height hh above the ground. It travels 59h\frac{5}{9}h of the total height in the last 1 s1\text{ s}. The height hh is: (use g=10 m/s2g=10\text{ m/s}^2)

A

5 m

B

25 m

C

45 m

D

58 m

Step-by-Step Solution

  1. Define Variables: Let tt be the total time taken to fall through height hh. Let u=0u=0 (starts from rest) and a=ga=g.
  2. Total Distance Equation: Using the second equation of motion : h=12gt2h = \frac{1}{2}gt^2
  3. Distance in (t-1) seconds: The object travels 59h\frac{5}{9}h in the last second. Therefore, the distance traveled in the first (t−1)(t-1) seconds is: h′=h−59h=49hh' = h - \frac{5}{9}h = \frac{4}{9}h Using the kinematic equation for this duration: 49h=12g(t−1)2\frac{4}{9}h = \frac{1}{2}g(t-1)^2
  4. Solve for Time (t): Substitute h=12gt2h = \frac{1}{2}gt^2 into the second equation: 49(12gt2)=12g(t−1)2\frac{4}{9}\left(\frac{1}{2}gt^2\right) = \frac{1}{2}g(t-1)^2 Divide both sides by 12g\frac{1}{2}g: 49t2=(t−1)2\frac{4}{9}t^2 = (t-1)^2 Take the square root of both sides: 23t=t−1or23t=−(t−1)\frac{2}{3}t = t - 1 \quad \text{or} \quad \frac{2}{3}t = -(t-1) Solving 23t=t−1\frac{2}{3}t = t - 1 gives 1=t−23t⇒1=13t⇒t=3 s1 = t - \frac{2}{3}t \Rightarrow 1 = \frac{1}{3}t \Rightarrow t = 3\text{ s}. (The other solution yields t<1t < 1, which is invalid for the 'last 1 second' condition).
  5. Calculate Height (h): Substitute t=3 st=3\text{ s} back into the height equation: h=12(10)(3)2=5×9=45 mh = \frac{1}{2}(10)(3)^2 = 5 \times 9 = 45\text{ m}
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