Question

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 parameters

i) The max. Height to which the stone will rise in its flight

ii) 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 1

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.