Question

a stone is thrown upward to a height of 10 meter .After what time will it fall onto the Earth and what height can be reached by the stone if its initial velocity is double​

Answer 1

$\sf\red{Answer}$

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$\sf\green{Given - }$

$\bf h = 10m$

$\bf a = g = \underline + 9.8 m/s^2.$

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$\sf\green{To \:find - }$

1. Total time i.e. Time taken to reach the highest point + time taken to reach ground from max. height.

2. Height reached by stone if initial velocity is doubled.

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$\sf\red{Solution - }$

1.

First case -

In the first case when object reached the maximum height time taken equals -

$\longrightarrow$$\bf a = g = -9.8 m/s^2$ (because object is going upwards)

$\longrightarrow$$\bf v = 0 m/s$(because object reached max. height.)

$\longrightarrow$$\bf h = 10 m$

To find time first we have to find initial velocity -

By using the third equation of motion -

$\bf\red{v^2 = u^2 + 2as }$

$\implies$$\bf\pink{0 = u^2 - 2 \times 10 \times 9.8}$

$\implies$$\bf\pink{196 = u^2 }$

$\implies$$\bf\pink{u = 14 m/s }$

Initial velocity of stone = 14 m/s

Time to reach maximum height -

$\longrightarrow$$\bf u = 14 m/s$

$\longrightarrow$$\bf v = 0 m/s$

$\longrightarrow$$\bf a = -9.8 m/s^2$

By using the first equation of motion -

$\bf\red{v = u + at }$

$\implies$$\bf\pink{0 = 14 - 9.8t }$

$\implies$$\bf\pink{9.8t = 14 }$

$\implies$$\bf\pink{t = 14/9.8 }$

$\implies$$\bf\pink{t = 1.4 sec }$

Time taken by stone to maximun height = 1.4 sec

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Second case -

Time taken by object to reach the ground from max. height to ground -

in this case the initial velocity will be equal to zero and final velocity will be equal to initial velocity found in first case.

$\longrightarrow$$\bf u = 0 m/s$

$\longrightarrow$$\bf v = 14 m/s$

$\longrightarrow$$\bf a = 9.8 m/s^2$

By using the first equation of motion -

$\bf\red{v = u + at }$

$\implies$$\bf\pink{14 = 0 + 9.8t }$

$\implies$$\bf\pink{9.8t = 14 }$

$\implies$$\bf\pink{t = 14/9.8 }$

$\implies$$\bf\pink{t = 1.4 sec }$

Time to reach ground = 1.4 sec

Total time = 1.4 + 1.4 = 2.8 sec

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2.

Max. Height reached by stone if initial velocity is doubled -

Initial velocity in the first case = 14 m/s , Intital velocity is doubled = 14 × 2 = 28m/s

$\longrightarrow$$\bf u = 28m/s$

$\longrightarrow$$\bf v = 0 m/s$

$\longrightarrow$$\bf a = -9.8 m/s^2$

By using the third equation of motion -

$\bf\red{v^2 = u^2 + 2as }$

$\implies$$\bf\pink{0 = 28^2 - 2 \times s\times 9.8}$

$\implies$$\bf\pink{784 = 19.6s}$

$\implies$$\bf\pink{s = 40m }$

If the initial velocity is doubled , the hieght reached by stone = 40m

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Answer 2

$\underline {\underline{{\purple{ \sf Given }}}}$

• Height (h) ➠ 10 m
• Acceleration due to gravity (g) ➠ -9.8 N
• Final Velocity (v) ➠ 0 m/s

$\underline {\underline{{ \purple{\sf To \: Find }}}}$

• The velocity at which it was thrown upward
• Time taken by the object to reach the highest point
• If Initial Velocity is doubled, then it's height

$\underline {\underline{{ \green{\sf Calculating \: Initial \: Velocity \: (u) }}}}$

Formula Used :- $\underline{\underline{\boxed{ \gray{\sf{v}^{2} - {u}^{2} = 2gh }}}}$

Substituting Values

☞ (0)² - u² = 2 × -9.8 × 10

☞ 0 - u² = 2 × - 98

☞ -u² = -196

☞ u² = 196

☞ u = √196

☞ u = 14

${\green{ \underbrace{\boxed{\underline{\underline{\purple{ \tt \therefore Initial \: Velocity = 14 \: m/s }}}}}}}$

Now,

We have got u, so we can calculate time taken to reach maximum point.

Formula Used :- $\underline{\underline{\boxed{ \pink{\sf v = u + gt}}}}$

Substituting Values

☞ 0 = 14 + -9.8 × t

☞ -14 = -9.8t

☞ 14 = 9.8t

☞ t = 14/9.8

☞ t = 1.43

Note :- The Body was going upward so value of 9.8 will be in negative

${\gray{ \underbrace{\boxed{\underline{\underline{\blue{ \tt \therefore Time \: taken \: to \: reach \: highest \: point \: is \: 1.43 \: seconds }}}}}}}$

$\underline {\underline{{ \green{\sf Calculating \: Total \: Time }}}}$

☞ 1.43 + 1.43 → 2.86

Now, we will calculate, the height when, initial velocity is doubled

$\underline {\underline{{ \green{\sf Calculating \: Height \: (h) }}}}$

Initial Velocity ➠ 14 × 2 ➠ 27

Formula Used :- $\underline{\underline{\boxed{ \pink{\sf{v}^{2} - {u}^{2} = 2gh }}}}$

Substituting Values

☞ (0)² - (28)² = 2 × -9.8 × h

☞ 0 - 774 = -19.6h

☞ 774 = 19.6h

☞ h = 774/19.6

☞ h = 40

${\blue{ \underbrace{\boxed{\underline{\underline{\orange{ \tt \therefore Initial \: Velocity \: if \: speed \: is \: doubled \: will \: be \: 40 \: m }}}}}}}$

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