Question

A particle is dropped under gravity ( $ g = 9.8m/{s^2} $ ) from a height h and it travels a distance $ \dfrac{{9h}}{{25}} $ in the last second the height ‘h’ is:
A) $ 100m $
B) $ 125{\text{ }}m $
C) $ 145{\text{ }}m $
D) $ 167.5{\text{ }}m $

Answer 1

Hint: It is very important to have a good knowledge of laws of motion and the three equations of motion. Another important point to note is that when the particle was dropped it was at rest which means that the initial velocity $ u = 0m/s $ . Further care should be taken while putting the signs of different values.

Complete step by step answer:
Let us first note down the expressions which are given to us,
Distance in last second $ D = \dfrac{{9h}}{{25}} $
 $ g = 9.8m/{s^2} $
Since when the particle was dropped it was at rest so, Initial velocity $ u = 0m/s $
Height $ h = ? $
By developing his three laws of motion, Newton revolutionized science.
An object at rest remains at rest, and an object in motion remains in motion at constant speed and in a straight line unless acted on by an unbalanced force.
The acceleration of an object depends on the mass of the object and the amount of force applied.
Whenever one object exerts a force on another object, the second object exerts an equal and opposite on the first.
The equations of motion are as follows:
 $ 1)v = u + at $
 $ 2)s = ut + \dfrac{1}{2}a{t^2} $
 $ 3){v^2} - {u^2} = 2as $
Let us consider that a particle takes time $t$ to reach the ground.
Using equation of motion
The total distance will be $ $
 $ s = ut + \dfrac{1}{2}g{t^2} $
 $ h = 0 + \dfrac{1}{2} \times 9.8 \times {t^2}......(1) $
The particle covers the distance D' in (t-1 )second,
 $ D' = h - D $
 $ D' = h - \dfrac{{9h}}{{25}} $
Now after further solving, we get,
 $ D' = \dfrac{{16h}}{{25}} $
Now by again using the equation of motion, we get,
 $ \dfrac{{16h}}{{25}} = 0 + \dfrac{1}{2}g{(t - 1)^2}......(2) $
Now by dividing equation (2) by equation (1),
 $ \dfrac{{16}}{{25}} = \dfrac{{{{(t - 1)}^2}}}{{{t^2}}} $
After cross multiplying, we get ,
 $ \dfrac{{16}}{{25}} \times {t^2} = {(t - 1)^2} $
 $ \dfrac{4}{5} \times t = (t - 1) $
 $ t = 5\sec $
Now putting the value of t in equation (1), we get,
 $ h = \dfrac{1}{2} \times 9.8 \times 5 \times 5 $
 $ h = 122.5m $
Hence the height from which the particle was dropped from is 122.5 m.

Additional Information:
Sir Isaac Newton PRS was an English mathematician, physicist, astronomer, theologian, and author who is widely recognised as one of the greatest mathematicians and most influential scientists of all time. Sir Isaac Newton worked in many areas of mathematics and physics. He developed the theories of gravitation in $ 1666 $ when he was only $ 23 $ years old. In $ 1686 $ , he presented his three laws of motion in the Principia Mathematica Philosophiae Naturalis.

Note:
Be careful while applying the conditions given in the question. While solving the question, keep in mind the distance travelled by the particle in the last second. Calculation mistakes are possible, so try to avoid them. We must know the method to the second equation of motion in order to solve the problem.