Question

A stone is thrown in a vertically upward direction with a velocity o with an innecoal velocity of 40 m/s find maximum height reached by stone

Answer 1

u=40m/s

v=0m/s (kyuki ek point pr uper aake wo rest me ho gya tha)

a=g=-10m/s (kyuki retardation hui thi )

$then \\ t = \frac{v - u}{a}$

because

$a = \frac{v - u}{t }$

t=0-40/-10

=4min

$s = ut + \frac{1}{2} a {t}^{2}$

=40×4 +1/2×(-10 ×4×4)

= 160+1/2×(-160)

=160+(-80)

=80 m

ye jo aaya h 80 m wo distance/height h.

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

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

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

⠀

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