Question

\blue{\bold{\underline{\underline{Answer:}}}}
Answer:
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\green{\therefore{\text{Maximum\:height=80\:m}}}∴Maximumheight=80m

\green{\therefore{\text{Net\:displacement=0}}}∴Netdisplacement=0

\green{\therefore{\text{Total\:distance\:covered=160m}}}∴Totaldistancecovered=160m

\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}
Step−by−stepexplanation:
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\begin{lgathered}\green{ \underline \bold{Given :}} \\ : \implies \text{Initial \: velocity(u) = 40 \: m/s} \\ \\ : \implies \text{Final \: velocity(v) = 0 \: m/s} \\ \\ : \implies \text{Acceleration (a)= - 10 \: m}/{s}^{2} \\ \\ \red{ \underline \bold{To \: Find :}} \\ : \implies \text{Maximum \: height = ?} \\ \\ : \implies \text{Net \:displacement = ?} \\ \\ : \implies \text{Total \: distance \: covered= ?}\end{lgathered}
Given:
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:⟹Initial  velocity(u) = 40  m/s
:⟹Final  velocity(v) = 0  m/s
:⟹Acceleration (a)= - 10  m/s
2

ToFind:
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:⟹Maximum  height = ?
:⟹Net displacement = ?
:⟹Total  distance  covered= ?
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• According to given question :

\begin{lgathered}\bold{Using \: second \: equation \: of \: motion } \\ : \implies {v}^{2} = {u}^{2} + 2as \\ \\ : \implies {0}^{2} = {40}^{2} + 2 \times ( - 10) \times s \\ \\ : \implies 0 = 1600 - 20 \times s \\ \\ : \implies - 1600 = - 20 \times s \\ \\ : \implies \frac{ - 1600}{ - 20} = s \\ \\ \green{ : \implies \text{s = 80 \: m}} \\ \\ \bold{For \: Net \: displacement } \\ : \implies Net \: displacement =( final - initial)position \\ \\ : \implies Net \: displacement = 80 - 80 \\ \\ \green{ : \implies \text{Net \: displacement = 0 }}\\ \\ \text{Because \: final \: position \: and \: initial } \\ \text{position \: of \: stone \: is \: same \: so} \\ \text{net \: displacement \: is \: zero} \\ \\ \bold{For \: Total \: distance \: covered} \\ : \implies Distance = 2 \times max \: height \\ \\ : \implies Distance = 2 \times 80 \\ \\ \green{ : \implies \text{Distance = 160 \: m}}\end{lgathered}
Usingsecondequationofmotion
:⟹v
2
=u
2
+2as
:⟹0
2
=40
2
+2×(−10)×s
:⟹0=1600−20×s
:⟹−1600=−20×s
:⟹
−20
−1600
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=s
:⟹s = 80  m
ForNetdisplacement
:⟹Netdisplacement=(final−initial)position
:⟹Netdisplacement=80−80
:⟹Net  displacement = 0 
Because  final  position  and  initial 
position  of  stone  is  same  so
net  displacement  is  zero
ForTotaldistancecovered
:⟹Distance=2×maxheight
:⟹Distance=2×80
:⟹Distance = 160  m
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Answer 1

Answer:

what do you want to ask in it