Question

A stone is thrown vertically upward with an initial velocity of 40m/s. taking g=10m/s .find the maximum height reached by the stone. what is the net displacement and the total distance covered by the stone?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Maximum\:height=80\:m}}}$

$\green{\therefore{\text{Net\:displacement=0}}}$

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

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given :}} \\ : \implies \text{Initial \: velocity(u) = 40 \: m/s} \\ \\ : \implies \text{Final \: velocity(v) = 0 \: m/s} \\ \\ : \implies \text{Acceleration (a)= - 10 \: m}/{s}^{2} \\ \\ \red{ \underline \bold{To \: Find :}} \\ : \implies \text{Maximum \: height = ?} \\ \\ : \implies \text{Net \:displacement = ?} \\ \\ : \implies \text{Total \: distance \: covered= ?}$

• According to given question :

$\bold{Using \: second \: equation \: of \: motion } \\ : \implies {v}^{2} = {u}^{2} + 2as \\ \\ : \implies {0}^{2} = {40}^{2} + 2 \times ( - 10) \times s \\ \\ : \implies 0 = 1600 - 20 \times s \\ \\ : \implies - 1600 = - 20 \times s \\ \\ : \implies \frac{ - 1600}{ - 20} = s \\ \\ \green{ : \implies \text{s = 80 \: m}} \\ \\ \bold{For \: Net \: displacement } \\ : \implies Net \: displacement =( final - initial)position \\ \\ : \implies Net \: displacement = 80 - 80 \\ \\ \green{ : \implies \text{Net \: displacement = 0 }}\\ \\ \text{Because \: final \: position \: and \: initial } \\ \text{position \: of \: stone \: is \: same \: so} \\ \text{net \: displacement \: is \: zero} \\ \\ \bold{For \: Total \: distance \: covered} \\ : \implies Distance = 2 \times max \: height \\ \\ : \implies Distance = 2 \times 80 \\ \\ \green{ : \implies \text{Distance = 160 \: m}}$

Answer 2

Given :--

• initial velocity = 40m/s.
• g = 10m/s.

To Find :---

• Maximum Height Reached by the stone ?
• Net Displacement ?
• Total Distance Covered by the stone ?

Concept And Formula used :---

→ when the stone is thrown upward we take acceleration due to gravity as negative ..

→ As the stone reaches the maximum Height, Means its final velocity will be zero..

→ Third Equation of motion :- v² - u² = 2as { Here , v is final velocity, u is initial velocity, a is the acceleration and s is displacement } .

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Solution :----

→ u = 40m/s .

→ g = (-10) m/s² { as stone is thrown upward} .

→ v = 0 { at maximum Height}

Putting values in Above Told Formula we get :--

→ (0)² - (40)² = 2 * (-10) * h

→ (-1600) = (-20)h

Dividing both sides by (-20) we get,

→ h = 80 m ...

So, the maximum Height at which the stone can reached is 80m..

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(2) Total Distance covered by stone now = 2 × maximum Height

= 2 × 80

= 160m.

Hence, the stone will cover 160m distance .

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(3) Now, as the stone comes back at its initial Position ,, So the Displacement of Stone will be = 0 .

Hence, displacement will be Zero.

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