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
[tex]s_{y}=ut-\dfrac{1}{2}gt^2+h_{0}[/tex]
Where, [tex]h_{0}[/tex] =initial height
u = initial velocity
g = acceleration due to gravity
[tex]s_{y}[/tex] = vertical height
Here, [tex]s_{y}[/tex] is zero at ground.
Put the value into the equation
[tex]0=0-\dfrac{1}{2}\times9.8t^2+10[/tex]
[tex]4.9t^2=10[/tex]
[tex]t=\sqrt{\dfrac{10}{4.9}}[/tex]
[tex]t = 1.42\ sec[/tex]
Hence, The container to reach the ground at 1.42 sec.
