Back to Directory
NEET PHYSICSMedium

A body falling from a high Minaret travels 40 m40 \text{ m} in the last 2 s2 \text{ s} of its fall to the ground. The height of the Minaret in meters is: (take g=10 m/s2g=10 \text{ m/s}^2)

A

60

B

45

C

80

D

50

Step-by-Step Solution

  1. Define Variables: Let the total height of the Minaret be HH and the total time taken to reach the ground be tt. The body falls from rest, so initial velocity u=0u = 0.
  2. Total Distance Equation: Using the second equation of motion s=ut+12at2s = ut + \frac{1}{2}at^2 with a=g=10 m/s2a = g = 10 \text{ m/s}^2: H=12gt2=5t2…(1)H = \frac{1}{2}gt^2 = 5t^2 \quad \dots(1)
  3. Distance Before Last 2 Seconds: The distance covered in (t−2)(t - 2) seconds is: H′=12g(t−2)2=5(t−2)2H' = \frac{1}{2}g(t - 2)^2 = 5(t - 2)^2
  4. Apply Condition: The distance traveled in the last 2 seconds is the difference between the total height and the height covered in (t−2)(t-2) seconds. H−H′=40H - H' = 40 5t2−5(t−2)2=405t^2 - 5(t - 2)^2 = 40 5[t2−(t2−4t+4)]=405[t^2 - (t^2 - 4t + 4)] = 40 5(4t−4)=405(4t - 4) = 40 20(t−1)=40  ⟹  t−1=2  ⟹  t=3 s20(t - 1) = 40 \implies t - 1 = 2 \implies t = 3 \text{ s}
  5. Calculate Height: Substitute t=3 st = 3 \text{ s} back into equation (1): H=5(3)2=5(9)=45 mH = 5(3)^2 = 5(9) = 45 \text{ m} .
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started