Question

a body is release from a top of a bulding of height 19.6m find its velocity which the body hits the ground​

Answer 1

$\mathcal{ANSWER:-}$

Known Terms:-

u = intial velocity

v = final velocity

s = distance

a = acceleration

t = Time

Given:-

s = 19.6m

u = 0

(since, the object starts from rest)

a = 9.8 ${ms^{-2}}$

To Find:-

The final velocity (v)

Formula Used :-

$\boxed{v^{2} - u^{2} = 2as}$

Solution:-

v² - 0² = 2×9.8×19.6

v² = 384.16

v = √384.16

v = 19.6 ${ms^{-1}}$

$\mathfrak{Hence,\:the\:velocity\:of\:the\:object}$

$\mathfrak{when\:it\:hits\:the\:ground\:is\:19.6ms^{-1}}$

Answer 2

1. initial u=0
2. final v=?
3. height=19.6
4. g=9.8m^2

v^2-u^2-2gh

v^2-0^2=2×9.8×19.6

v^2=19.6 ×19.6

v^2=(19.6)^2

v^2=384.16

v=

$\sqrt{384.16}$

v=19.6ms