Question

A stone is thrown vertically upward with an initial velocity of 20 m/s. Find the maximum height it reaches and the time taken by it to reach the height.

Answer 1

Given:-

• Initial Velocity = 20m/s

• Acceleration due to Gravity = -10m/s

• Final Velocity = 0m/s ( At maximum Height , v = 0)

To Find:-

• The Maximum Height it reaches and the time taken by it reach the height.

Formulae used:-

• S = ut + ½ × a × t²

• v = u + at

Where,

S = Height

u = Initial Velocity

t = Time

a = Acceleration

v = Final Velocity

Now,

→ v = u + at

→ v = 20 + -10 × t

→ 0 = 20 - 10t

→ - 20 = -10t

→ t = 2s.

Therefore,

→ S = ut + ½ × a × t²

→ S = 20 × 2 + ½ × -10 × (2)²

→ S = 40 + -5 × 4

→ S = 40 - 20

→ S = 20m.

Hence, The maximum Height reached by stone is 20m and in 2s.

Answer 2

Explanation:

$\setlength{\unitlength}{1mm}\begin{picture}(8,2)\thicklines\multiput(9,1.5)(1,0){50}{\line(1,2){2}}\multiput(35,7)(0,4){13}{\line(0,1){2}}\put(10.5,6){\line(3,0){50}}\put(35,60){\circle*{10}}\put(37,7){\large\sf{u = 20m/s}}\put(37,55){\large\sf{v = 0 m/s}}\put(20,61){\large\textsf{\textbf{Stone}}}\end{picture}$

• Initial Velocity ( u ) = 20 m/s

• Final Velocity ( v ) = 0 m/s

• Gravity = - 10 m/s²

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$\underline{\bigstar\:\textsf{By Third Equation of Gravity :}}$

$:\implies\sf v^2-u^2=2gh\\\\\\:\implies\sf (0)^2-(20)^2=2 \times (-10) \times h\\\\\\:\implies\sf 0- 400=-20 \times h\\\\\\:\implies\sf -400 = - 20 \times h\\\\\\:\implies\sf \dfrac{ - 400}{ - 20} = h\\\\\\:\implies\underline{\boxed{\sf h = 20 \:metres}}$

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