Question

a stone is dropped from a tower of height 3600cm. find the velocity by which it strikes the ground. Also find the time taken to reach the ground.
please I need its answer​

Answer 1

★ Given :-

• initial velocity of the stone = 0

(as the stone is dropped )

• acceleration due to gravity = +10 m/s

( as the motion of the stone is acting along gravity )

• height of the tower = 3600 cm

We need to convert it into m in order to that simply divide by 100

height of tower = 36 m

★ To find :-

• time taken to reach the ground
• velocity with which it strikes the ground

★ Solution :-

As the body moves uniformly under the influence of gravity we can use the equations of motion in order to solve this question

By using second equation of motion ,

$\implies\boxed{\mathtt{s = ut + \dfrac{1}{2} g {t}^{2} }}$

here ,

• s = distance
• u = initial velocity
• t = time taken
• g= acceleration due to gravity

$\implies\mathtt{s = \dfrac{1}{2} g {t}^{2} }$

$\implies\mathtt{ {t}^{2} = \dfrac{2s}{g} }$

$\implies\mathtt{ {t}^{2} = \dfrac{2 \times 36}{10} }$

$\implies\mathtt{ {t}^{2} =7.2}$

$\implies \boxed{\mathtt{ t =2.68 \: sec}}$

The time taken by the stone to reach ground is 2.68 sec

_____________

Now , in order to find final velocity let's use the first equation of motion ,

$\implies\boxed{\mathtt{v = u + gt}}$

here ,

• v = final velocity
• u = initial velocity
• g = acceleration
• t = time taken

$\implies\mathtt{v = gt}$

Now let's substitute the given values in the above equation ,

$\implies\mathtt{v = 2.68 \times 10}$

$\implies \boxed{\mathtt{v = 26.8 \dfrac{m}{s} }}$

The velocity of the stone just before it strikes the ground is 26.8 m/s

Answer 2

Answer:

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Explanation:

Given :-

initial velocity of the stone = 0

(as the stone is dropped )

acceleration due to gravity = +10 m/s

( as the motion of the stone is acting along gravity )

height of the tower = 3600 cm

We need to convert it into m in order to that simply divide by 100

height of tower = 36 m

★ To find :-

time taken to reach the ground

velocity with which it strikes the ground

★ Solution :-

As the body moves uniformly under the influence of gravity we can use the equations of motion in order to solve this question

By using second equation of motion ,

\implies\boxed{\mathtt{s = ut + \dfrac{1}{2} g {t}^{2} }}⟹

s=ut+

2

1

gt

2

here ,

s = distance

u = initial velocity

t = time taken

g= acceleration due to gravity

\implies\mathtt{s = \dfrac{1}{2} g {t}^{2} }⟹s=

2

1

gt

2

\implies\mathtt{ {t}^{2} = \dfrac{2s}{g} }⟹t

2

=

g

2s

\implies\mathtt{ {t}^{2} = \dfrac{2 \times 36}{10} }⟹t

2

=

10

2×36

\implies\mathtt{ {t}^{2} =7.2}⟹t

2

=7.2

\implies \boxed{\mathtt{ t =2.68 \: sec}}⟹

t=2.68sec

The time taken by the stone to reach ground is 2.68 sec