Answer:
The container to reach the ground at 1.42 sec.
Explanation:
Given that,
Height h = 10 m
Velocity = 1 m/s
Using equation of motion for vertical height
\(s_{y}=ut-\dfrac{1}{2}gt^2+h_{0}\)
Where, \(h_{0}\) =initial height
u = initial velocity
g = acceleration due to gravity
\(s_{y}\) = vertical height
Here, \(s_{y}\) is zero at ground.
Put the value into the equation
\(0=0-\dfrac{1}{2}\times9.8t^2+10\)
\(4.9t^2=10\)
\(t=\sqrt{\dfrac{10}{4.9}}\)
\(t = 1.42\ sec\)
Hence, The container to reach the ground at 1.42 sec.
