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Derivation of Position-Velocity Relation by Graphical Method

A ball projec...

Question

A ball projected vertically upwards from A,the top of the tower,reaches the ground in t1 seconds. If it is projected vertically downwards from A with the same velocity,it reaches the ground in t2 seconds.if it falls freely from A the time taken to reach the ground is?

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Solution

Suppose h is the height of the tower and u is the projection velocity.
and suppose the required time is 't'.
For upward projection,
$h = - {ut}_{1} + \frac{1}{2} {gt}_{1}^{2} - - - - - - (i) For downward projection, h = {ut}_{2} + \frac{1}{2} {gt}_{2}^{2} - - - - - - - (ii) For the free fall, h = 0 + \frac{1}{2} g t^{2} - - - - - - - - (iii) Multiplying (i) by t_{2} and (ii) by t_{1} h t_{2} = - {ut}_{1} t_{2} + \frac{1}{2} {gt}_{1}^{2} t_{2} {ht}_{1} = {ut}_{2} t_{1} + \frac{1}{2} {gt}_{2}^{2} t_{1} Now adding, h (t_{1} + t_{2}) = \frac{1}{2} g t_{1} t_{2} (t_{1} + t_{2}) or h = \frac{1}{2} g t_{1} t_{2} Putting the above value in (iii), \frac{1}{2} g t_{1} t_{2} = \frac{1}{2} g t^{2} or t = \sqrt{t_{1} t_{2}}$

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