First slide

Rectilinear Motion

Question

In the last second of free fall, a body covered $\frac{3}{4} t h$ of its total path. Then the height from which the body is released will be $(g = 10 m / s^{2})$

Moderate

A

10m
B

20m
C

30m
D

40m

Solution

If ‘n’ is the last second, then the distance travelled by a freely falling body in ‘n’ sec is
$\begin{array}{l} s = \frac{1}{2} g t^{2} (∵ u = 0) \\ = \frac{1}{2} g n^{2} \dots \dots . . (1) and s_{n} = g (n - \frac{1}{2}) \dots \dots . . . (2) \\ Given that s_{n} = \frac{3}{4} s \\ \Rightarrow g (n - \frac{1}{2}) = \frac{3}{4} (\frac{1}{2} g n^{2}) \\ ∴ (2 n - 1) = \frac{3}{4} (n^{2}) \\ \Rightarrow \frac{(2 n - 1)}{n^{2}} = \frac{3}{4} \\ \Rightarrow n = 2 s \end{array}$
The height from which the body is released is $s = \frac{1}{2} g n^{2}$
$\begin{array}{l} = \frac{1}{2} \times 10 \times (2)^{2} \\ = 20 m \end{array}$

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