Question

A player throws a ball upwards with the initial speed of 29.4m/s . Taking gravity=9.8m/s the time taken by the ball to return to the player's hand is​

Answer 1

Given:

• Initial velocity of ball = 29.4 m/s
• Magnitude of Acceleration due to gravity taken = 9.8 m/s²

To find:

• Time taken by ball to return player's hand, t =?

Formula required:

• First equation of motion

     v = u + a t

• Second equation of motion

    s = u t + 1/2 a t²

• Third equation of motion

     2 a s = v² - u²

[ where v is final velocity, u is initial velocity, a is acceleration, t is time taken and s is distance covered ]

Solution:

Let, time taken by ball to reach the maximum height be t₁ and maximum height ball covered be s,

then, final velocity of ball at highest point will be zero; taking initial velocity of ball = 29.4 m/s; and acceleration due to gravity = -9.8 m/s²

Using first equation of motion

→ v = u + a t

→ 0 = ( 29.4 ) + ( -9.8 ) t₁

→ -29.4 = -9.8  t₁

→ t₁ = 29.4 / 9.8

→ t₁ = 3 sec

then,

Using Third equation of motion

→ 2 a s = v² - u²

→ 2 ( -9.8 ) s = ( 0 )² - ( 29.4 )²

→ - 19.6  s = -864.36

→ s = 864.36 / 19.6

→ s = 44.1  m

Now,  

Let time taken by ball to reach the player's hand again be t₂ , then taking initial velocity 0, distance covered by ball 44.1 m, and acceleration due to gravity 9.8 m/s².

Using second equation of motion

→ s = u t + 1/2 a t²

→ 44.1 = ( 0 ) ( t₂ ) + 1/2 ( 9.8 ) t₂²

→ 44.1 = 4.8  t₂²

→ t₂² = 9.1875

→ t₂ = 3.03 sec

Now,

total time taken by the ball to return to the hands of player is

→ t = t₁ + t₂

→ t = 3 + 3.03

→ t = 6.03 second [ approx. ]

Therefore,

• total time taken by the ball to return to the hands of player is 6.03 seconds. (approximately).

Answer 2

Answer:

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

• Initial Velocity ( u ) = 29.4 m/s
• Final Velocity ( v ) = 0 m/s [ As Ball will be at rest after attending maximum height ]
• Acceleration due to Gravity = - 9.8 m/s²

⠀

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

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

$\rule{180}{1.5}$

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

As no external forces are acting on Ball, Hence

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

⠀

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