Question

Question 5.
A stone is dropped from the top of a 40 m high tower. Calculate its speed after 2 s. Also find the speed with
which the stone strikes the ground.
Answer:
​

Answer 1

$\huge{\underline{\pink{\tt{Given,}}}}$

• A stone is dropped from the top of a 40 m(height) high tower.
• Time(t) = 2 sec.
• Initial Velocity(u) = 0
• Gravitational Acceleration(g) = 9.8 m/s²

$\huge{\underline{\pink{\tt{To\:Find,}}}}$

• Find the speed with  which the stone strikes the ground.

$\huge{\underline{\pink{\tt{Formula\:Used:}}}}$

• $\bigstar \boxed{\sf{v = u + at}}$
• $\bigstar \boxed{\sf{v^{2} = u^{2} + 2gs}}$

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

According to the First Condition :-

• Speed After 2 Seconds.

$\longrightarrow \underline{\sf{We\:know\:that,}}$

$\bigstar \boxed{\sf{v = u + at}}$

Now, Put the Value in this Formula :-

$\longmapsto \sf{v = 0 + 9.8 \times 2}$

$\longmapsto \boxed{\sf{v = 19.6\:m/s}}$

Therefore,

• The Speed of stone after 2 second will be 19.6 m/s.

$\underline{\red{\sf{Now,We\:will\:find\:the\:speed\:with\:which\:the\:stone\:strikes\:the\:ground :-}}}$

∴We know that :-

$\bigstar \boxed{\sf{v^{2} = u^{2} + gs}}$

$\longmapsto \sf{v^{2} = 0 + 2 \times 9.8 \times 40}$

$\longmapsto \sf{v^{2} = 784}$

$\longmapsto \sf{v = \sqrt{784}}$

$\longmapsto \boxed{\sf{v = 28\:m/s}}$

Therefore,

• Speed with which the stone strikes the ground is 28 m/s.

$\rule{200}2$

Answer 2

$\huge\bf{\underline{\underline{\sf{\red{Answer:}}}}}$

$\large\implies{\boxed{\sf{\blue{The\:speed\:of\:the\:stone\:after\:2s\:is\:19.6 m/s.}}}}$

$\large\implies{\boxed{\sf{\orange{The\:speed\:with\:which\:the\:stone\:strikes\:the\:ground\:is\:28\:m/s}}}}$

$\huge\bf{\underline{\underline{\sf{\purple{Explanation:}}}}}$

✤ Given :

• Height = 40m
• Time = 2 sec

✤ To Find :

• It's speed after 2s.
• The speed with which the stone strikes the ground.

✤ Solution :

Formulae used :

➤$\large{\boxed{\sf{\pink{v=u+at}}}}$

where, u = 0 and a = g(acceleration due to gravity)

Substituting the values, we get,

:$\Rightarrow{\sf{v = 0+9.8 \times 2}}$

:$\Rightarrow{\sf{v = 19.6m/s}}$

The speed of the stone after 2s is 19.6 m/s.

After that, we have to find the stone strikes the ground,

Fomulae used :

➤$\large{\boxed{\sf{\pink{v^{2}=u^{2}+2gs}}}}$

Substituting the values, we get,

:$\Rightarrow\displaystyle{\sf{v^{2} = 0+2 \times 9.8 \times 40}}$

:$\Rightarrow\displaystyle{\sf{v^{2}=784}}$

:$\Rightarrow\displaystyle{\sf{v^{2}=784}}$

:$\Rightarrow\displaystyle{\sf{v=\sqrt{784}}}$

:$\Rightarrow\displaystyle{\sf{v=28m/s}}$

Therefore, the speed with which the stone strikes the ground is 28m/s.

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