Question

A body is thrown vertically upwards from $A$ , the top of a tower. It reaches the ground in time $t_{1} .$ If it is thrown vertically downwards from $A$ with the same speed it reaches the ground in time $t_{2} .$ If it is allowed to fall freely from $A$ , then find the time it takes to reach the ground.

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Accepted Answer

Suppose the body be projected vertically upwards from $A$ with a speed $u_{0}$ . Using equation $A$
For first case $- h = u_{0} t_{1} - (\frac{1}{2}) g t_{1}^{2} . . . . (i)$

For second case, $- h = - u_{0} t_{2} - (\frac{1}{2}) g t_{2}^{2} . . . . (ii)$

From $(i)$ $- (ii) \Rightarrow 0 = u_{0} (t_{2} + t_{1}) + (\frac{1}{2}) g (t_{2}^{2} - t_{1}^{2})$ $\Rightarrow u_{0} = (\frac{1}{2}) g (t_{1} - t_{2}) . . . . (iii)$

Put the value of $u_{0}$ in (ii), we get, $- h = - (\frac{1}{2}) g (t_{1} - t_{2}) t_{2} - (\frac{1}{2}) g t_{2}^{2}$

$\Rightarrow h = \frac{1}{2} g t_{1} t_{2} . . . . (iv)$

For third case, $u = 0, t = ?$ , we get

$- h = u t - \frac{1}{2} g t^{2} \Rightarrow h = (\frac{1}{2}) g t^{2} . . . . (v)$

Combining $(iv)$ and $(v)$ , we get
$\frac{1}{2} g t^{2} = \frac{1}{2} g t_{1} t_{2} or t = \sqrt{t_{1} t_{2}}$

B M Sharma Solutions for Chapter: Kinematics I, Exercise 30: ILLUSTRATION 4.28

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 30: ILLUSTRATION 4.28

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 30: ILLUSTRATION 4.28 with Hints & Solutions

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