Question

A stone is dropped freely from the top of a tower and it reaches the ground in 6 sec taking (g = 10ms2). Calculate the height of the Tower

​

Answer 1

Given :

• Time (t) = 6 seconds
• Initial velocity (u) = 0 m/s
• Acceleration of gravity (g) = 10 m/s²

To find :

• Height of the tower

According to the question,

${\boxed{ \sf{s = ut + \dfrac{1}{2} {gt}^{2} }}} \\ \\ : \implies\sf{s = 0 \times 6 + \dfrac{1}{2} \times 10 \times {6}^{2} } \\ \\ : \implies \sf{s = 0 + \frac{1}{2} \times 10 \times 36 } \\ \\ : \implies \sf{s = 5 \times 36} \\ \\ { \underline { \boxed{ \sf{ : \implies s = 180 \: m }}}}$

So,the height is 180 m..

More formula :

$\bullet \: \sf{v = u + gt} \\ \\ \bullet \: \sf{ {v}^{2} = {u}^{2} + 2gs} \\ \\ \bullet \: \sf{s = ut + \dfrac{1}{2} {gt}^{2} }$

• u stand for Initial velocity
• v stands for Final velocity
• t stands for Time
• g stands for Acceleration of gravity
• s stands for Distance

Answer 2

Given:-

• Time taken to reach Earth = 6 seconds [t]
• $g$ = 10 m/s²
• (Initial velocity = 0 m/s - u)

To Find:-

Height of the tower (= distance covered by the stone)

We know,

$h = ut + \frac{1}{2}gt^2$

where,

• $h$ = Height,
• $u$ = Initial velocity,
• $t$ = Time taken &
• $g$ = Acceleration due to gravity.

$\therefore h = \frac{1}{2}gt^2$

$\implies h = \frac{1}{2} (10 m/s^2)(6s)^2$

$\implies h = (5 m/s^2)(36 s^2)$

$\implies h = (5 m)(36)$

$\boxed{ \implies h = 180 m}$