Question

A stone dropped from the roof of a
building takes 4s to reach the ground.
Calculate the height of the building and its
velocity
take g=10 m/s2​

Answer 1

Answer:

nitial velocity = 0

time taken = 4s

acceleration = g = 9.8m/s^2

using 2nd equation of motion  

s= ut +(1/2)at^2

= (0)(4) +(1/2)(9.8)(4)(4)

= 78.4

Explanation:

Answer 2

★ Provided Question ★

A stone dropped from the roof of a building takes 4s to reach the ground.Calculate the height of the building and its velocity.(take g=10 m/s^2)

How to solve?

Here, in the question we are given that the stone is dropped from the height so it's initial velocity will be 0m/s.And it takes 4s to reach the ground. Acceleration due to gravity is 10m/s^2 , acceleration will be here positive as it is dropped from a height.

• We will use here the formula || h= ut + $\sf {\dfrac {1}{2}g{t}{2}}$|| to calculate the height.
• To calculate its final velocity we will use the formula || v = u + gt ||.
• At last we'll get our required answer.

★ Required Solution ★

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

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

• Time taken (t) = 4s

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

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

• Height (h)
• final velocity (v)

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

We know that:

• ${\boxed {\huge {\bf {\pink { h =ut + \dfrac{1}{2}g{t}^{2}}}}}}$

Substitute the values:

$\sf { : \implies h =ut + \dfrac{1}{2}a{t}^{2}}$

$\sf { : \implies h =(0 \times 4) + \dfrac{1}{\cancel{2}} \times \cancel{10} \times {4}^{2}}$

$\sf { : \implies h =0 + 1 \times 5 \times 16 }$

$\sf { : \implies h = 1 \times 80 }$

$\sf \red { : \implies h = 80m }$

Now, calculating velocity:

$\sf \pink { Initial \: velocity = 0m/s }$

Whenever the object is dropped from a height,initial velocity (u) is 0.Therefore, stone's initial velocity will be 0m/s.

$\sf \pink { Final \: velocity = 40m/{s}^{2} }$

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

Substituting values:

$\sf { : \implies v = 0 + 10 \times 4 }$

$\sf { : \implies v = 0+ 40 }$

$\sf \red { : \implies v = 40m/{s}^{2} }$

Therefore, final velocity of ball is $\sf { 40m/{s}^{2} }$

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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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