First slide

Rectilinear Motion

Question

A stone is dropped from the top of a tower. When it has fallen by 5 m from the top, another stone is dropped from a point 25 m below the top. Ifboth stones reach the ground at the same moment, then height of the tower is (take g = 10 m/s²)

Moderate

A

45 m
B

50 m
C

60 m
D

65 m

Solution

Velocity of I^st stone when passing at A
$\begin{array}{l} V^{2} = 0 + 2.10.5 \Rightarrow V = 10 m / s \\ and S_{1} - S_{2} = 20 m \\ \Rightarrow (10 \cdot t + \frac{1}{2} 10 \cdot t^{2}) - (\frac{1}{2} \cdot 10 \cdot t^{2}) = 20 \\ At t = 2 s, S_{2} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times 2^{2} = 20 m \end{array}$
Hence, height of the tower,
$H = S_{1} + S_{2} = 25 + 20 = 45 m$

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