Question

a boy standing at the top of 20m height drop a stone.(take g= 10 m/s) calculate
a) the velocity with which it hits the ground
b)time taken by the stone to reach the ground
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Answer 1

★ Provided Question ★

A boy standing at the top of 20m height drop a stone.(take g= 10 m/s^2).Calculate:

• a) the velocity with which it hits the ground
• b)time taken by the stone to reach the ground

How to solve?

Here, in this question we are given that the height(h) of the building is 20m and the acceleration due to gravity (g) is 10m/s^2. The initial velocity will be 0m/s as it is dropped from the height.

• To calculate the velocity with which it hits the ground i.e its final velocity, we'll use the formula || $\sf { {v}^{2}-{u}^{2}= 2gh}$||

• To calculate the time, we'll use here the formula || $\sf { v=u+gt}$||

• So,let's solve!

★ Required Solution ★

${\underline {\underline {\rm { Given: } }}}$

• Initial velocity (u) = 0m/s [ As it dropped from the height. ]

• Height of the building (h) = 20m

• acceleration due to gravity (g) = $\sf { 10m/{s}^{2} }$

${\underline {\underline {\rm { To \: calculate: } }}}$

• final velocity (v)
• time taken by the stone to reach the ground (t)

${\underline {\underline {\rm { Calculation : } }}}$

We know that:

• ${\boxed {\huge {\bf {\pink { 2gh ={v}^{2}-{u}^{2}}}}}}$

Substituting values-

$\sf { :\implies 2 \times 10 \times 20 ={v}^{2}-{0}^{2} }$

$\sf { :\implies 20 \times 20 ={v}^{2}-0 }$

$\sf { :\implies 400 ={v}^{2}-0 }$

$\sf { :\implies 400 ={v}^{2} }$

$\sf { :\implies v= \sqrt{400} }$

$\sf \red { :\implies v= 20m/s }$

Therefore, final velocity of the stone is 20m/s.

Now calculating time taken:

We know that:

• ${\boxed {\huge {\bf {\pink { v=u+gt}}}}}$

Substituting values:

$\sf { :\implies 20 = 0 + 10 \times t}$

$\sf { :\implies 20 = 10 t}$

$\sf { :\implies t = \dfrac{20}{10} }$

$\sf \red { :\implies t = 2s }$

Therefore, time taken by the stone to reach the ground is 2 seconds.

Points to remember:

• Whenever we throw the object vertically upwards, then final velocity(v) is 0m/s.

• Whenever the object is dropped vertically from a height then initial velocity (u) is 0m/s.

• Whenever we throw the object vertically upwards, then acceleration due to gravity (g) is -10m/s^2.

• Whenever the object is dropped vertically from a height then acceleration due to gravity (g) is 10m/s^2.

Equations of motion:

• ${\boxed {\huge {\bf {\pink { v=u+at}}}}}$
• ${\boxed {\huge {\bf {\pink { s =ut + \dfrac{1}{2}a{t}^{2}}}}}}$
• ${\boxed {\huge {\bf {\pink { 2as ={v}^{2}-{u}^{2}}}}}}$

Equation of motion for freely falling bodies:

• ${\boxed {\huge {\bf {\pink { v=u+gt}}}}}$
• ${\boxed {\huge {\bf {\pink { h =ut + \dfrac{1}{2}g{t}^{2}}}}}}$
• ${\boxed {\huge {\bf {\pink { 2gh ={v}^{2}-{u}^{2}}}}}}$

Where,

v denotes final velocity.

u denotes initial velocity.

'a' or 'g' denotes acceleration.

's' or 'h' denotes distance/height

t denotes time.

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