Question

14. A stone is released from the top of a tower of height
19.6 m. Calculate its final velocity. RB
​

Answer 1

Explanation:

final velocity(v) =

$\large{ \sqrt{2gh} }$

where,

h is the height

putting the value,

$\large{ \sqrt{2 \times 9.8 \times 19.6} }$

= $\large{ \sqrt{19.6 \times 19.6} }$

= $\large \boxed{{ \sqrt{384.16} \: m{s}^{ - 2} }}$

Answer 2

$\huge{\bf{\underline{\overline{\mid{Answer}\mid}}}}$

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

• Height of the tower(h)= 19.6 m

• Initial velocity(u)= 0 m/s

• Acceleration (a)= 10 m/s²

$\huge{\underline{\tt{To\:Find:-}}}$

• Final velocity (v)

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

By using 3rd equation of motion

${ \boxed{ \bf{v {}^{2} = u {}^{2} + 2gh}}}$

putting the values, we get

${ \tt{v {}^{2} = 0 + 2 \times 10 \times 19.6}} \\ { \tt{v {}^{2} = 392}} \\ { \tt{ v = \sqrt{392} }} \\ { \boxed{ \bf{v = 14 \sqrt{2} }}}$

•°• the final velocity is 14√2 m/s.