Question

A body draps a ball from the top of tower of height 19.6cm.calculate velocity of ball just before touch the ground (g=10m/s²).

Answer 1

Explanation:

u = 0

v^2 = u^2 + 2*g*s

v^2 = 0 + 2*10*0.196

v^2 = 3.92

taking square root

v = 1.98 m/s

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

h = 19.6 cm = 19.6/100 m = 0.196 m.

u = 0 m/s.

g = 10 m/s².

$\huge{\bold{\underline{Explanation:-}}}$

Applying third kinematic equation.

$\large{\bold{v^2 - u^2 = 2as}}$

Substituting the values.

$\large{v^2 - 0 = 2 \times 10 \times 0.196}$

$\large{v^2 = 2 \times 1.96}$

$\large{v = \sqrt{ 2 \times 1.96}}$

$\large{v = 1.4 \sqrt{2}}$

$\large{v = 1.4 \times 1.4}$

As √2 = 1.4

Now,

$\huge{\boxed{\boxed{v = 1.96 \: m/s}}}$

So,the velocity before it touches the ground is 1.96 m/s.