First slide

Rectilinear Motion

Question

A body released from certain height above the ground describes $\frac{7}{16}$ of the total height in the last second of its fall. Then it is falling from a height equal to $(g = 9.8 m / s^{2})$

Moderate

A

78.4 m
B

72.4m
C

64.4 m
D

60.5 m

Solution

Let ‘n’ be the last second of its fall
the total height covered after ‘n’ sec is
$\begin{array}{l} h = u t + \frac{1}{2} g t^{2} \\ \Rightarrow h = 0 + \frac{1}{2} g n^{2} \\ = \frac{1}{2} g n^{2} \dots \dots \dots . . (1) \end{array}$
The distance travelled in the last second of its fall is $s_{n} = u + g (n - \frac{1}{2})$
$\begin{array}{l} \Rightarrow \frac{7}{16} h = 0 + g \frac{(2 n - 1)}{2} \\ \Rightarrow \frac{7}{16} (\frac{1}{2} g n^{2}) = g \frac{(2 n - 1)}{2} \\ \Rightarrow \frac{2 n - 1}{n^{2}} = \frac{7}{16} \end{array}$
On solving $n = 4 s$
$∴$ The time taken by the body $t = 4 s$
$∴$ The height $h = \frac{1}{2} g n^{2}$
$\begin{array}{l} = \frac{1}{2} \times 9.8 \times (4)^{2} \\ = 9.8 \times 8 \\ = 78.4 m \end{array}$

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