Question

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 1

$\huge\star{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

$\huge{\red{\ddot{\smile}}}{\huge{\pink{\ddot\smile}}}{\huge{\purple{\ddot{\smile}}}}{\huge{\orange{\ddot{\smile}}}}{\huge{\blue{\ddot{\smile}}}}$

Given that,

Height = 40 m

Time t =2 sec

We know that,

v=u+gt

v=0+9.8×2

v=19.6m/s

The speed of stone is 19.6 m/s

Now, the speed with which the stone strikes the ground

v=28m/s

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

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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