Question

A body is projected horizontally with a velocity of 20m/s from the top of a tower of height 490m.
The distance at which it falls from the top of the tower is​

Answer 1

$\sf{\underline{\underline{ \purple{ \large{Given:}}}}}$

✰ Initial velocity in vertical direction ( uₓ ) = 0 m/s

✰ Initial velocity in horizontal direction ( uᵧ ) = 20 m/s

✰ Vertical height or distance ( h ) = 490 m

$\\ \\ \\ \sf{\underline{\underline{ \purple{ \large{To \: Find:}}}}}$

✠ The distance at which body falls from the top of the tower i.e, horizontal distance ( S )

$\\ \\ \\ \sf{\underline{\underline{ \purple{ \large{Solution:}}}}}$

Let the vertical distance be h

ㅤㅤㅤㅤㅤ

Applying,

$\\ \\ \sf{ \blue{ \large{ \underline{ \boxed{ \sf{h = u_{x}t + \dfrac{1}{2} a {t}^{2} }}}}}}$

As the body is falling freely under the action of gravity,

$\sf{ \blue{ \large{ \underline{ \boxed{ \sf{h = u_{x}t + \dfrac{1}{2} g {t}^{2} }}}}}}$

Put the values in,

$\implies \tt{490 = (0)t + \dfrac{1}{2} \times 10 \times {t}^{2} }$

$\implies \tt{490 = 0 \times t+ \dfrac{1}{2} \times 10 \times {t}^{2} }$

$\implies \tt{490 = 0 + \dfrac{1}{2} \times 10 \times {t}^{2} }$

$\implies \tt{{t}^{2} = \dfrac{2 \times 490}{10} }$

$\implies \tt{{t}^{2} = \dfrac{2 \times 49 \cancel0}{ 1\cancel0} }$

$\implies \tt{{t}^{2} = 2 \times 49 }$

$\implies \tt{t = \sqrt{ 2 \times 49 } }$

$\implies \tt{t = 7 \sqrt{ 2 } \: s }$

Now for finding horizontal distance,

Applying,

$\\ \\ \sf{ \blue{ \large{ \underline{ \boxed{ \sf{S = u_{y}t + \dfrac{1}{2} a {t}^{2} }}}}}}$

$\\ \implies\tt{S = 20 \times 7 \sqrt{2} + \dfrac{1}{2} \times 0 \times {t}^{2}}$

$\implies\tt{S = 20 \times 7 \sqrt{2} }$

$\implies\tt{S = 140 \sqrt{2} }$

$\implies\tt{S \approx 197.99 \: m }$

$\sf{ \therefore The \: distance \: at \: which \: body \: falls \: from \: the \: top \: of \: the \: tower ( S ) \approx\sf{ \blue{ \large{ \underline{ \boxed{ \sf{197.99 \: m}}}}}}}$

─━━━━━━━━━──━━━━━━━━━─