Question

a body is thrown down from a tower of height 192 m with initial velocity 2 m/s find the time taken to hit the ground velocity just before it hits the ground. plz answer it's very urgent ..i'ill mark 1st correct answer as brainliest

Answer 1

Answer:

HERE IS YOUR ANSWER BUDDY IN THE ATTACHMENT

Answer 2

$\huge {\sf{\underline {\underline{ G iven : - }}}}$

• A body is thrown down from a height of 192 m with initial velocity 2m/s

$\huge {\sf {\underline {\underline{To \: find : - }}}}$

• Time taken by the body to hit the ground

• Velocity with which it hits the ground

$\huge {\sf {\underline {\underline{Solution : - }}}}$

We have ,

u = 2 m/s

h = 192 m

g = a = 9.8m/s²

From Third equation of motion,

$\large{\underline{\boxed{ \bigstar{ | {v}^{2} - {u}^{2} = 2as}}}}$

Where ,

• v is final velocity

• u is initial velocity

• a is acceleration

• s is distance covered or height

$\sf : \implies \: v {}^{2} - (2) {}^{2} = 2(9.8)(192) \\ \\ \sf : \implies \: {v}^{2} - 4 = 3763.2 \\ \\ \sf : \implies \: v {}^{2} = 3767.2 \\ \\ \sf : \implies \: v = \sqrt{3767.2} \\ \\ \sf : \implies {\bold {\boxed {\pink {\sf{v = 61.37 \: m {s}^{ - 1} }}}}}$

From Second equation of motion ,

$\large {\underline{\bold {\boxed{ \bigstar { \: | v = u + a t }}}}}$

Where ,

• u is initial velocity

• a is acceleration

• v is final velocity

• t is time

$\sf: \implies \: 61.37 = 2 + (9.8)(t) \\ \\ \sf : \implies 59.37 = 9.8t \\ \\ \sf : \implies \: t = \frac{59.37}{9.8} \\ \\ \sf : \implies {\bold {\boxed {\pink{ t = 6.24 \: s}}}}$

∴ The velocity with which it hits the ground and the time taken to hit the ground are 61.37 m/s and 6.24 sec