Question

A stone falls from the top of a building and travels 53.9 m in the last second before it reaches the ground. Find the height of the ground?
Class 11 physics
...No s*tupid answers please​

Answer 1

Question

A stone falls from the top of a building and travels 53.9 m in the last second before it reaches the ground. Find the height of the ground?

given

initial velocity= 0

distance travelled= 53.9 metre

accleration= gravity = 9.8 m/sec²

To find

height of the building

Solution

˙❥According to the equations of kinematics-

$\boxed{ \to \: \sf { \color{green}S_{nth} = u + \frac{a}{2} + (2n - 1)}}$

where

• u denotes inital Velocity.
• a denotes Acceleration.
• n denotes time.

❄️Substitute the values in the above equation of kinematics-

$\sf{ \to \: 53.9 = 0 + \frac{9.8}{2} (2n - 1)} \\ \\ \sf { \to \: 53.9 \times 2 = 9.8(2n - 1)} \\ \\ \sf { \to \: 107.8 = 19.6n - 9.8} \\ \\ \sf{ \to \: n = \frac{117.6}{19.6} } \\ \\ \sf \color{blue}{\to \: n = 6 \: seconds}$

According to second equation of kinematics-

$\boxed{ \color{magenta}\sf{ \to \: s = ut + \frac{1}{2} a {t}^{2} }}$

where

• s denotes height of the building
• u denotes the inital Velocity
• a denotes accleration
• t denotes time

$\sf{ \to \: s = 0 + \frac{1}{2} \times 9.8 \times ( {6})^{2} } \\ \\ \sf{\to \: s = \frac{9 .8 \times 36}{2} } \\ \\ \sf \color{gold}{ \to \: s = 176.4 \: metre}$

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Answer 2

Answer:

ǫᴜᴇsᴛɪᴏɴ:-

A stone falls from the top of a building and travels 53.9 m in the last second before it reaches the ground. Find the height of the ground?

ɢɪven:-

initial velocity= 0

distance travelled= 53.9 metre

accleration= gravity = 9.8 m/sec²

To find:-

height of the building

Solution

˙❥According to the equations of kinematics-

$\boxed{ \to \: \sf { \color{green}S_{nth} = u + \frac{a}{2} + (2n - 1)}}$

where

u denotes inital Velocity.

a denotes Acceleration.

n denotes time.

❄️Substitute the values in the above equation of kinematics-

$\begin{gathered} \sf{ \to \: 53.9 = 0 + \frac{9.8}{2} (2n - 1)} \\ \\ \sf { \to \: 53.9 \times 2 = 9.8(2n - 1)} \\ \\ \sf { \to \: 107.8 = 19.6n - 9.8} \\ \\ \sf{ \to \: n = \frac{117.6}{19.6} } \\ \\ \sf \color{blue}{\to \: n = 6 \: seconds}\end{gathered}$

According to second equation of kinematics-

$\boxed{ \color{magenta}\sf{ \to \: s = ut + \frac{1}{2} a {t}^{2} }}$

$\begin{gathered} \sf{ \to \: s = 0 + \frac{1}{2} \times 9.8 \times ( {6})^{2} } \\ \\ \sf{\to \: s = \frac{9 .8 \times 36}{2} } \\ \\ \sf \color{gold}{ \to \: s = 176.4 \: metre}\end{gathered}$

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