Question

A particle is dropped under gravity from rest from a height $h(g=9.8\frac{m}{se{c}^2})$ and it travels a distance $\frac{9h}{25}$ in the last second, the height h is

• A. 100 m
• B. 122.5 m
• C. 145 m
• D. 167.5 m

Answer 1

The correct option is B 122.5 m

The object starts from the rest

Distance travelled in last second= $\frac{9h}{25}$

Let n is the total time

$h=0+\frac{1}{2}g{n}^2$......................1

Distance travelled in (n-1) second = $h-\frac{9h}{25}=\frac{16h}{25}$

$\frac{16h}{25}=\frac{1}{2}g(n-1{)}^2$

by using value of h from eq 1 in eq 2

$\frac{16}{25}g{n}^2=\frac{1}{2}g{(n-1)}^2$

taking under root both side

$\frac{4}{5}n=n-1$

$\frac{n}{5}=1$

$n=5$

Now,

$h=\frac{1}{2}\times 9.8\times 5^2$

$=122.5m$