Question

A ball iS thrown vertically upwards with a velocity of 13 ms^-1 calculate the time taken to the to reach maximum height

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Time\:taken=1.3\:sec}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Initial \: velocity(u) = 13 \: m/s \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Time \: taken \: to \: reach \: maximum \: height(t) =?$

• According to given question :

$\tt \circ \: Final \: velocity = 0\:m/s\\ \\ \tt\circ\: Acceleration = - 10 { \: m/s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 0 = 13 + ( - 10) \times t \\ \\ \tt: \implies 0 = 13 - 10 \times t \\ \\ \tt: \implies - 13 = - 10 \times t \\ \\ \tt: \implies t = \frac{ - 13}{ - 10} \\ \\ \tt: \implies t = \frac{13}{10} \\ \\ \green{\tt: \implies t = 1.3 \: sec} \\ \\ \green{\tt \therefore Time \: taken \: to \: reach \: maximum \: height \: is \: 1.3 \: sec}$

Answer 2

Answer:

⋆ DIAGRAM

$\setlength{\unitlength}{1.5mm}\begin{picture}(60,30)\thicklines\put(10,10){\line(1,0){39}}\put(30,15){\vector(0,2){22}}\put(25,12){\bf Ground}\put(30,40){\circle*{20}}\put(33,40){\bf Ball}\put(31,15.8){\sf u = 13 m/s}\put(31,35.8){\sf v = 0 m/s}\put(15,25){\sf a = - 9.8m/s ^\sf2 }\put(12.6,22){\sf(Against Gravity)}\multiput(10,10)(1,0){40}{\line(-2,-5){1.5}}\end{picture}$

$\rule{150}{1}$

$\frak{Given}\begin{cases}\textsf{Initial Velocity (u) = 13 m/s}\\\textsf{Final Velocity (v) = 0 m/s}\\\textsf{Acceleration (g) = - 9.8 m/s \sf^2 }\\\textsf{Time (t) = ?}\end{cases}$

$\underline{\bigstar\:\textbf{Let's Consider a Formula :}}$

$:\implies\sf v=u+gt\\\\\\:\implies\sf 0=13+(-9.8) \times t\\\\\\:\implies\sf 0=13-9.8t\\\\\\:\implies\sf 9.8t=13\\\\\\:\implies\sf t=\dfrac{13}{9.8}\\\\\\:\implies\underline{\boxed{\sf t=1.32\: seconds}}$

⠀

$\therefore\:\underline{\textsf{Time taken to reach max ^\text n height is \textbf{1.32 seconds}}}.$