First slide

Rectilinear Motion

Question

A body is thrown vertically upwards from A, the top of a tower. It reaches the ground in time t₁. lf it is thrown vertically downwards from A with the same speed, it reaches the ground in time t₂. If it is allowed to fall freely from A, then the time it takes to reach the ground is given by

Difficult

A

$t = \frac{t_{1} + t_{2}}{2}$
B

$t = \frac{t_{1} - t_{2}}{2}$
C

$t = \sqrt{t_{1} t_{2}}$
D

$t = \sqrt{\frac{t_{1}}{t_{2}}}$

Solution

Suppose the body be projected vertically upwards from A with a speed u₀.
$\begin{array}{l} Using equation s = u t + (\frac{1}{2}) a t^{2}, we get \\ For first case: - h = u_{0} t_{1} - (\frac{1}{2}) g t_{1}^{2} . . . . . . (i) \\ For second case: - h = - u_{0} t_{2} - (\frac{1}{2}) g t_{2}^{2} . . . . . . (i i) \end{array}$
From (i) and (ii) we get
$h = \frac{1}{2} g t_{t} t_{2}$ ……(iii)
$For third case: u = 0, t = ?$
$- h = 0 \times t - (\frac{1}{2}) g t^{2} or h = (\frac{1}{2}) g t^{2}$ …..(iv)
Combining Eq. (iii) and Eq. (iv), we get
$\frac{1}{2} g t^{2} = \frac{1}{2} g t_{1} t_{2} or t = \sqrt{t_{1} t_{2}}$

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