Question

a stone is released from top of tower of height 19.6m .Caculate the final velocity just before touching the ground ​

Answer 1

Solution

Stone is a free falling body

From the Question,

• Height of the tower,t = 19.6m

• Initial Velocity of the body is zero as it starts from rest,u = 0 m/s

According to Sign Convention,

• Acceleration,a = g = 9.8m/s²

Using the Kinematic Equation,

${\sf{v {}^{2} - u {}^{2} = 2gh}} \\ \\ \longrightarrow \boxed{{\sf{v {}^{2} = 2gh }}}$

Substituting the values,we get:

v² = 2(9.8)(19.6)

→v² = (19.6)²

→v = 19.6 m/s

The velocity of the object before it touches the ground is 19.6m/s

Answer 2

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

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

S = H = 19.6 m

u = 0 m/s

a = g = 9.8 m/s²

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

As "H" ,"u" ,"g" are given so we should apply third kinematic equation.

From third kinematic equation.

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

Substituting the values.

${ {v}^{2} - {0}^{2} = 2 \times 9.8 \times 19.6}$

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

${v = 19.6 m/s}$

$\huge{\boxed{\boxed{v = 19.6 m/s}}}$

so,the final velocity before touching the ground is 19.6 m/s.