2. A stone is thrown vertically upward with a velocity 20m/s from the

1.	2. A stone is thrown vertically upward with a velocity 20m/s from the top of a tower 25m high. Make calculations for the following parametersi) The max. Height to which the stone will rise in its flightii) Velocity of the stone during its downward travel at a point in the same level as the point of projection.iii) Time required for the stone to reach the ground.
Answer» Explanation : here, u = Initial Velocity = 20m/s Finial Velocity = v = 0m/s a = - 10m/s² We took 'a' negative as it is GOING against GRAVITATIONAL pull. For Calculating : $\large \rm \: g = 10 \dfrac{m}{ {s}^{2} } \\$ Formula : 2as = v² - u² Transposing The Terms $\large \tt : \implies \: s = \dfrac{ {v}^{2} - {u}^{2} }{2a} \\ \\ \large \tt : \implies \: s = \dfrac{ 0 - {20}^{2} }{ - 2 \times 10} m\\ \\ \\ \large \tt : \implies \: s = \dfrac{ - 400 }{ - 20} \\ \\ \\ \large \tt : \implies \: s = \dfrac{ \cancel - 400 }{ \cancel - 20} \\ \\ \\ \large \tt : \implies \: s = \dfrac{ \cancel{400} \: \: 20 }{ \cancel{20}} \\ \\ \\ \large \tt \boxed{ \sf\therefore\: s = 20m } \\ \\$ Time Taken : $\\ \large \rm \leadsto \: t = \frac{v - u}{a} \\ \\$ $\large \sf : \implies \: t = \dfrac{0 - 20}{ - 10} \\ \\ \\ \large \sf : \implies \: t = \dfrac{- 20}{ - 10} \\ \\ \\ \large \sf \therefore \underline{ \underline{\: t = 2seconds}} \longrightarrow \: t_1 \\ \\$ It reach 20m froma a HEIGHT of 25m Finding Distance $\\ \large \sf \: s = 20 + 25 = 45m \\ \\$ Since, Object started falling from 45m So Initial Velocity = u = 0 Acceleration = - 10m/s² = Finding Time Taken : Using Formula $\\ \large \rm \leadsto \: s = ut + \frac{1}{2} a {t}^{2} \\ \\$ $\large \tt : \implies \: 45 = 0 \times t + \dfrac{1}{2} \times 10 \times {t}^{2} \\ \\ \\ \large \tt : \implies \: {t}^{2} = 45 \times \frac{2}{10} \\ \\ \\ \large \tt : \implies \: {t}^{2} = \frac{90}{10} \\ \\ \\ \large \tt : \implies \: {t}^{2} = 9 \\ \\ \large \tt : \implies \: {t}^{} = \sqrt{ 9} \\ \\ \large \tt \boxed{ \underline{ \sf \therefore \: t = 3seconds}} \longrightarrow \: t_2 \\$ For TOTAL Time : Adding : $\\ \large \sf \: t = \: t_1 + \: t_2 \\ \\ \large \sf \implies \: t = \: 2 + 3 \\ \\ \large \sf \leadsto \boxed{ \rm t = 5 \: seconds} \\$ So, the stone will reach the ground in 5 seconds after being thrown upwards from tower.

2. A stone is thrown vertically upward with a velocity 20m/s from the top of a tower 25m high. Make calculations for the following parametersi) The max. Height to which the stone will rise in its flightii) Velocity of the stone during its downward travel at a point in the same level as the point of projection.iii) Time required for the stone to reach the ground.

Answer»

Explanation :

here,

u = Initial Velocity = 20m/s

Finial Velocity = v = 0m/s

a = - 10m/s²

We took 'a' negative as it is GOING against GRAVITATIONAL pull.

For Calculating :

$\large \rm \: g = 10 \dfrac{m}{ {s}^{2} } \\$

Formula :

2as = v² - u²

Transposing The Terms

$\large \tt : \implies \: s = \dfrac{ {v}^{2} - {u}^{2} }{2a} \\ \\ \large \tt : \implies \: s = \dfrac{ 0 - {20}^{2} }{ - 2 \times 10} m\\ \\ \\ \large \tt : \implies \: s = \dfrac{ - 400 }{ - 20} \\ \\ \\ \large \tt : \implies \: s = \dfrac{ \cancel - 400 }{ \cancel - 20} \\ \\ \\ \large \tt : \implies \: s = \dfrac{ \cancel{400} \: \: 20 }{ \cancel{20}} \\ \\ \\ \large \tt \boxed{ \sf\therefore\: s = 20m } \\ \\$

Time Taken :

$\\ \large \rm \leadsto \: t = \frac{v - u}{a} \\ \\$

$\large \sf : \implies \: t = \dfrac{0 - 20}{ - 10} \\ \\ \\ \large \sf : \implies \: t = \dfrac{- 20}{ - 10} \\ \\ \\ \large \sf \therefore \underline{ \underline{\: t = 2seconds}} \longrightarrow \: t_1 \\ \\$

It reach 20m froma a HEIGHT of 25m

Finding Distance

$\\ \large \sf \: s = 20 + 25 = 45m \\ \\$

Since,

Object started falling from 45m

So Initial Velocity = u = 0

Acceleration = - 10m/s² =

Finding Time Taken :

Using Formula

$\\ \large \rm \leadsto \: s = ut + \frac{1}{2} a {t}^{2} \\ \\$

$\large \tt : \implies \: 45 = 0 \times t + \dfrac{1}{2} \times 10 \times {t}^{2} \\ \\ \\ \large \tt : \implies \: {t}^{2} = 45 \times \frac{2}{10} \\ \\ \\ \large \tt : \implies \: {t}^{2} = \frac{90}{10} \\ \\ \\ \large \tt : \implies \: {t}^{2} = 9 \\ \\ \large \tt : \implies \: {t}^{} = \sqrt{ 9} \\ \\ \large \tt \boxed{ \underline{ \sf \therefore \: t = 3seconds}} \longrightarrow \: t_2 \\$

For TOTAL Time :

Adding :

$\\ \large \sf \: t = \: t_1 + \: t_2 \\ \\ \large \sf \implies \: t = \: 2 + 3 \\ \\ \large \sf \leadsto \boxed{ \rm t = 5 \: seconds} \\$

So,

the stone will reach the ground in 5 seconds after being thrown upwards from tower.

Explanation :

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