Question

A body is vertically up from the top of a tower with 19.6m/s takes 5sec to reach the ground. Then height of the tow
er is

Answer 1

$\sf{\green{Given}}\begin{cases}</p><p>\sf{Initial \: Velocity\: (u) = 19.6 \:m/s}\\</p><p>\sf{Time \: taken \: (t) = 5 \: s}</p><p>\end{cases}$

$</p><p>\sf{\green{To\: find}}\\</p><p>\sf{\star\: \: Height \: of \: the \: tower \: (h) }</p><p>$

$\sf{\green{Solution}}$

Let the displacement be 'h'.

Here, considering the downward direction of acceleration to be the positive

$\large{\sf{\blue{\underline{Using \: equation \: of\: motion}}}}$

$\boxed{\large{\sf{\red{s = ut + \frac{1}{2}at^{2}}}}}$

$</p><p>\sf{\implies\: h = ut + \frac{1}{2}gt^{2}}\\</p><p>$

$</p><p>\sf{\implies\: h = 19.6 × 5 + \frac{1}{2} × 9.8 × 5^{2}}\\</p><p>$

$</p><p>\sf{\implies\: h = 19.6 × 5 + 4.9 × 25}\\</p><p>$

$</p><p>\sf{\implies\: h = 98 + 122.5}\\</p><p>$

$</p><p>\boxed{\sf{\implies\: s = 220.5\: m \: \: \green{(Answer)}}}\\ \\</p><p></p><p>\large{\sf{\green{\therefore\: Height\:of\:the\:tower = 220.5\:m}}}</p><p>$

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Note:

$\sf{\purple{\star\: \: a\: (upward) = g\: (-ve)}} \\</p><p>\sf{\purple{\star\: \: a\: (downward) = g\: (+ve)}} \\ \\</p><p></p><p>\sf{\red{\star\: \: g = 9.8\: m/s^{2}}}</p><p>$