Question

a person throws a ball upwards with an initial speed of 29.4m/s. To what height does the ball rise and after how long does the ball return to the players hand

Answer 1

Answer:

Ball rises to the height of 44.1m

Explanation:

initial speed (u)=29.4m/s

we know that $v^{2}$-$u^{2}$=2gh

ball reaches its maximum height then its comes to rest

so its final speed (v)=0m/s

we should take g=9.8m/$s^{2}$

now substitute these values in this

-(29.4 x 29.4)=2(9.8)h

if cancellation is done we get value of h is -44.1m

here we get -44.1m. this is because ball is coming in opposite direction

so we can leave the -.so the answer is 44.1m

Answer 2

Answer:

m:

$\setlength{\unitlength}{1mm}\begin{picture}(8,2)\thicklines\multiput(9,1.5)(1,0){50}{\line(1,2){2}}\multiput(33,7)(0,4){12}{\line(0,1){2}}\multiput(36,10)(0,4){12}{\line(0,1){2}}\put(10.5,6){\line(3,0){50}}\put(34,60){\circle*{10}}\put(8,8){\large\sf{u = 29.4 m/s}}\put(37,55){\large\sf{v = 0 m/s}}\put(22,61){\large\textsf{\textbf{Ball}}}\put(36,12){\vector(0, - 4){5}}\put(33,50){\vector(0,4){5}}\end{picture}$

⠀

$\underline{\bigstar\:\textsf{Using First Motion under Gravity :}}$

$:\implies\sf v = u + gt\\\\\\:\implies\sf 0=29.4\:m/s+(-\:9.8\:m/s^2) \times t\\\\\\:\implies\sf - \:29.4\:m/s = -\:9.8\:m/s^2 \times t\\\\\\:\implies\sf \dfrac{ - \:29.4\:m/s}{-\:9.8\:m/s^2} = t\\\\\\:\implies\underline{\boxed{\sf t = 3 \:seconds}}$

$\rule{180}{1.5}$

$\underline{\bigstar\:\textsf{According to the given Question :}}$

As no external forces are acting on Ball, Hence

$\dashrightarrow\textsf{Time of ascent = Time of decent}\\\\\\\dashrightarrow\sf Total\:Time=Time \:of \:ascent+Time \:of\:decent\\\\\\\dashrightarrow\sf Total\:Time=Time \:of \:ascent + Time \:of \:ascent\\\\\\\dashrightarrow\sf Total\:Time=3 \:seconds + 3 \:seconds\\\\\\\dashrightarrow \underline{ \boxed{\sf Total\:Time=6\:seconds}}$

⠀

$\therefore\:\underline{\textsf{Ball will return to players hand after \textbf{6 seconds}}}.$