First slide

Rectilinear Motion

Question

A ball is thrown vertically upward from the roof of a building with a certain velocity. It reaches the ground in 9 s. When it is thrown downward from the roof with the
same initial speed, it takes 4 s to come at ground. How much time (in second) will it take to reach at ground if it just released from the rest from the roof?

Moderate

Solution

Let the height of building is h.
$\begin{array}{l} Event I, - h = v_{0} t_{1} - \frac{1}{2} g t_{1}^{2} . . . . . . . . . . (i) \\ Event II, h = v_{0} t_{2} + \frac{1}{2} g t_{2}^{2} . . . . . . . . . . . . (i i) \\ Event III, h = \frac{1}{2} g t^{2} . . . . . . . . . . . (i i i) \end{array}$
On multiplying Eq. (i) by t₂ and Eq. (ii) by t₁ and add, we get
$\begin{array}{l} h (t_{1} + t_{2}) = \frac{1}{2} g t_{1} t_{2} (t_{1} + t_{2}) \\ Or \frac{1}{2} g t^{2} = \frac{1}{2} g t_{1} t_{2} \end{array}$
$∴ t = \sqrt{t_{1} \times t_{2}} = \sqrt{9 \times 4} = 6 s$

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