Question

Q18. An object is dropped from rest at a height of 100 m and simultaneously another object is dropped from
rest at a height 50 m. What is the difference in their heights after 2 s if both the objects drop with same
accelerations? How does the difference in heights vary with time?

Answer 1

$\large \clubs \:  \bf Given :  -$

• An object is dropped from rest from a height of 100 m .

• Simultaneously another object is dropped from rest from a height of 50 m.

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$\large \clubs \:  \bf To \: Find :  -$

• Difference in their heights after 2 s.

• How does the difference in heights vary with time ?

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$\large \clubs \:  \bf Solution :  -$

Here Acceleration is the Acceleration due to gravity (g).

Acceleration = a = g = 10 m/s²

We Have 2nd Equation of Kinematics as :

$\red \bigstar \: \boxed{\orange{ \boxed{ \bf{s =ut} + \dfrac{1}{2}{at}^{2}}}}$

★ For First Body :-

• Time = t = 2 second

• Initial Velocity = u = 0 m/s

• Acceleration = a = 10 m/s²

✏ Using 2nd Equation of Kinematics :

★ Distance covered by 1st body :

$\text s_1 = 0 + \frac{1}{2} \times 10 \times {2}^{2}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf s_1 = 20 \: m} }}}$

As Initial Height first body is 100 m

Hence,

Height of 1st body after 2s = 100 - 20

⇝Height of 1st body after 2s = 80 m

Now,

★ For Second Body :-

• Time = t = 2 second

• Initial Velocity = u = 0 m/s

• Acceleration = a = 10 m/s²

✏ Using 2nd Equation of Kinematics :

★ Distance covered by 2nd body :

$\text s_2 = 0 + \frac{1}{2} \times 10 \times {2}^{2}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf s_2 = 20 \: m} }}}$

As Initial Height Second body is 50 m

Hence,

Height of 2nd body after 2s = 50 - 20

⇝Height of 2nd body after 2s = 30 m

So,

Difference in their Heights after 2s = 80 - 30

$\underline{\bf \pink{\underline{Difference \: in \: their \: Heights}}}\\ \underline{\bf \pink{\underline{after \: 2s = 50 m }}}$

Also,

It concluded that initial difference between their Heights is equals to final difference between their Heights

So,

$\underline{\bf \pink{\underline{Difference\: in\: their\: Heights}}}\\ \underline{\bf \pink{\underline{ does\: not\: vary\: with\: time}}}$

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Answer 2

Given that , An object is dropped from rest at a height of 100 m and simultaneously another object is dropped from rest at a height 50 m.

Need To Find : The Difference in there height after 2 sec & How does the difference in heights vary with time ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀▪︎ We know that , to calculate their Distance covered in 2 sec we can use Second Equation of Motion .

》 Second Equation of motion is Given by :

$\qquad \qquad \star \:\:\underline {\boxed {\pmb{\sf \: s \:=\:ut \:+\:\bigg\lgroup \dfrac{1}{2} \bigg\rgroup \:gt^2\:}}}\:$

⠀Where ,

• s is the Distance Travelled,
• t is the Time Taken ,
• u is the Initial Velocity &
• g is the Acceleration due to Gravity .

$\qquad \qquad \bigstar \:\:\underline {\pmb{\pink { \sf \: Height \:\:of \:First \: Body \:after \:2 \:sec\:\::\:}}}$

$\qquad \dashrightarrow \sf \: s \:=\:ut \:+\:\bigg\lgroup \dfrac{1}{2} \bigg\rgroup \:gt^2 \:$

$\qquad \dashrightarrow \sf \: s \:=\:\big( 0 \big) 2 \:+\:\bigg\lgroup \dfrac{1}{2} \bigg\rgroup \:g\:\times\:\big(2\big)^2 \:$

$\qquad \dashrightarrow \sf \: s \:=\:\big( 0 \big) 2 \:+\:\bigg\lgroup \dfrac{1}{2} \bigg\rgroup \:\big(10\big)\:\times\:\big(2\big)^2 \:\:\qquad \because \: \bigg\lgroup \sf{ Acc^n \:_{\:(Gravity \:)}\:=\:10\:m/s^2 }\bigg\rgroup$

$\qquad \dashrightarrow \sf \: s \:=\:\big( 0 \big) 2 \:+\:\bigg\lgroup \dfrac{1}{2} \bigg\rgroup \:\big(10\big)\:\times\:\big(2\big)^2$

$\qquad \dashrightarrow \sf \: s \:=\: \:10\:\times\:2$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\purple { \: s\:\: \:=\: \:20 \:\:m\:}}}}}\:\:\bigstar$

$\qquad \dag\:\underline {\frak{ Height \:\:After \:2 \:sec\:\big( \:Initial \:Height \:=\:100\:m\:\big)\:\::}}$

$\qquad \dashrightarrow \sf \: Height \:_{1}\:\:=\: \:Initial \:Height \: \:- \:\:Height \:after\:2 \: sec\:$

$\qquad \dashrightarrow \sf \: Height \:_{1}\:\:=\: \:100\: \:- \:\:20\:$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\purple { \: Height _1\:\: \:=\: \:80 \:\:m\:}}}}}\:\:\bigstar$

$\qquad \qquad \bigstar \:\:\underline {\pmb{\pink { \sf \: Height \:\:of \:Second \: Body \:after \:2 \:sec\:\::\:}}}$

$\qquad \dashrightarrow \sf \: s \:=\:ut \:+\:\bigg\lgroup \dfrac{1}{2} \bigg\rgroup \:gt^2 \:$

$\qquad \dashrightarrow \sf \: s \:=\:\big( 0 \big) 2 \:+\:\bigg\lgroup \dfrac{1}{2} \bigg\rgroup \:g\:\times\:\big(2\big)^2 \:$

$\qquad \dashrightarrow \sf \: s \:=\:\big( 0 \big) 2 \:+\:\bigg\lgroup \dfrac{1}{2} \bigg\rgroup \:\big(10\big)\:\times\:\big(2\big)^2 \:\:\qquad \because \: \bigg\lgroup \sf{ Acc^n \:_{\:(Gravity \:)}\:=\:10\:m/s^2 }\bigg\rgroup$

$\qquad \dashrightarrow \sf \: s \:=\:\big( 0 \big) 2 \:+\:\bigg\lgroup \dfrac{1}{2} \bigg\rgroup \:\big(10\big)\:\times\:\big(2\big)^2$

$\qquad \dashrightarrow \sf \: s \:=\: \:10\:\times\:2$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\purple { \: s\:\: \:=\: \:20 \:\:m\:}}}}}\:\:\bigstar$

$\qquad \dag\:\underline {\frak{ Height \:\:After \:2 \:sec\:\big( \:Initial \:Height \:=\:50\:m\:\big)\:\::}}$

$\qquad \dashrightarrow \sf \: Height \:_{2}\:\:=\: \:Initial \:Height \: \:- \:\:Height \:after\:2 \: sec\:$

$\qquad \dashrightarrow \sf \: Height \:_{2}\:\:=\: \:50\: \:- \:\:20\:$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\purple { \: Height _2\:\: \:=\: \:30 \:\:m\:}}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \qquad \bigstar \:\:\underline {\pmb{\pink { \mathbb{ \:DIFFERENCE \:\:IN\:\:THEIR\:\: HEIGHT \:\: \:AFTER \:2 \:SEC\:\::\:}}}}$

$\\\qquad \dashrightarrow \sf Difference _{(\:Height \:)}\:=\: Height _1 \:-\:Height _2\:$

$\qquad \dashrightarrow \sf Difference _{(\:Height \:)}\:=\: 80 \:-\:30\:$

$\qquad \dashrightarrow \underline {\boxed {\pmb{\frak{\purple { \: Difference _{(\:Height \:)}\:\:\: \:=\: \:50 \:\:m\:}}}}}\:\:\bigstar$

⠀⠀⠀⠀ Hence, Difference in their Height after 2 sec will be 50 m .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⌬⠀Is the difference in heights vary with time ?

▪︎ As we , Can see that Here , Initial Difference between Height is 50 m [ i.e. 100 - 50 ] & Difference in Height after 2 sec is 50 m [ i.e. 80 - 30 = 50 ] , Here Initial Difference between Height and Height after 2 sec is Equal .

Therefore,

• Difference in their Height doesn't vary with time .