Question

a boy dropped a ball from the top of 20m building and starts the stopwatch . what will be the reading of stopwatch when ball will reach at ground surface, if the acceleration due to gravity is 10m/s^2?
plz answer it correctly.

Answer 1

Answer:

2 seconds

Explanation:

initial velocity(u) = 0m/s

accelaration(a) = g = 10m/s^{2}

distance(s) = height(h) = 20m

s = ut + 1/2at^{2}

h = ut + 1/2gt^{2}

20 = 0(t) + 1/2(10)t^{2}

20 = 5t^{2}

t^{2} = 20/5

t = \sqrt{4}

t = 2 seconds

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

s = 20m (Height)

a = 10 m/s² (Accele. due to gravity)

u = 0 (Initial velocity) [ Because Ball starts from rest]

Using 2nd Equation of motion :-

⇒ s = ut + 1/2 × at²

⇒ 20 = 0 × t + 1/2 × 10 × t²

⇒ 20 = 5 × t²

⇒ 20/5 = t²

⇒ t² = 4

⇒ t = √4

⇒ t = 2 seconds

Therefore,

Time taken by ball to reach earth's surface = 2 seconds

$\boxed{\begin{minipage}{11 cm} Additional Information \\ \\ \ Force=Mass\times Acceleration \\ \\ Kinetic Energy = \dfrac{1}{2}\times Mass\times (Velocity)^{2} \\ \\ Potential\; Energy=Mass\times Acceleration\;due\;to\;gravity\times Height \\ \\ Power = \dfrac{Total\;Work\;done}{Time}=\dfrac{Energy}{Time} \end{minipage}}$