Question

A stone is vertically thrown upwards with an initial velocity of 40 m/s^1 calculate maximum height reached​

Answer 1

Given:-

• Initial Velocity = 40m/s

• Final Velocity = 0 ( Maximum height )

• Acceleration due to gravity = -10m/s²

To Find:-

• The Maximum Height reached by Stone.

Formulae used:-

• v² - u² = 2gh

Where,

• v = Final Velocity
• u = Initial Velocity
• g = Acceleration due to gravity
• h = Height.

Now,

→ v² - u² = 2gh

→ (0) - (40)² = 2 × -10 × h

→ -1600 = -20h

→ 1600 = 20h

→ h = 1600/20

→ h = 80m

Hence, The maximum height reached by Stone is 80m.

SOME MORE EQUATION :-

• v = u + at

• s = ut + ½ × a × t²

• w = F × s

Answer 2

Answer:

$\setlength{\unitlength}{1mm}\begin{picture}(8,2)\thicklines\multiput(9,1.5)(1,0){50}{\line(1,2){2}}\multiput(35,7)(0,4){13}{\line(0,1){2}}\put(10.5,6){\line(3,0){50}}\put(35,60){\circle*{10}}\put(37,7){\large\sf{u = 40 m/s}}\put(37,55){\large\sf{v = 0 m/s}}\put(20,61){\large\textsf{\textbf{Stone}}}\end{picture}$

• Initial Velocity ( u ) = 40 m/s
• Final Velocity ( v ) = 0 m/s [As Stone will stop after attending maximum height]
• Acceleration due to Gravity = - 10 m/s²

⠀

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

$:\implies\sf v^2-u^2=2gh\\\\\\:\implies\sf (0)^2-(40)^2=2 \times (-10) \times h\\\\\\:\implies\sf 0- 1600=-20 \times h\\\\\\:\implies\sf - 1600 = - 20 \times h\\\\\\:\implies\sf \dfrac{ - 1600}{ - 20} = h\\\\\\:\implies\underline{\boxed{\sf h = 80 \:metres}}$

⠀

$\therefore\:\underline{\textsf{Max ^\textsf m height reached by stone is \textbf{80 metres}}}.$