Question

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Find the final velocity of the ball that is dropped from the second floor if the ball takes 18 seconds before hitting the ground. The acceleration due to gravity is g = 9.80 m/s²​​

Answer 1

Answer:

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Answer 2

Answer:

$\setlength{\unitlength}{1mm}\begin{picture}(8,2)\thicklines\multiput(9,1.5)(1,0){50}{\line(1,2){2}}\multiput(33,7)(0,4){12}{\line(0,1){2}}\multiput(36,10)(0,4){12}{\line(0,1){2}}\put(10.5,6){\line(3,0){50}}\put(34,60){\circle*{10}}\put(3,8){\large\sf{Time = 18 sec}}\put(37,55){\large\sf{v = 0 m/s}}\put(22,61){\large\textsf{\textbf{Ball}}}\put(36,12){\vector(0, - 4){5}}\put(33,50){\vector(0,4){5}}\end{picture}$

• Time taken = 18 seconds
• Initial velocity = 0 m/s
• Gravitational Acceleration = 9.8 m/s²
• Final Velocity = ?

$\displaystyle\underline{\bigstar\:\textsf{Using first Equation of motion :}}$

$\displaystyle\sf \dashrightarrow v = u+at$

• v = ?
• u = 0 m/s
• a = 9.8 m/s²
• t = 18 sec

$\\\displaystyle\sf \dashrightarrow v = 0+9.8\times 18$

$\displaystyle\sf \dashrightarrow v = 0+176.4$

$\displaystyle\sf \dashrightarrow\underline{\boxed{\sf v = 176.4 m/s}}$

$\displaystyle\therefore\:\underline{\textsf{Final Velocity of the ball is \textbf{ 176.4 m/s }}}$

Know More

• Second Equation of motion : s = ut+½at²

• Third equation of motion : v²-u² = 2as

Where,

1. s = Distance
2. u = Initial Velocity
3. v = Final Velocity
4. t = Time
5. a = Acceleration