Respuesta :
Answer:
The velocity of the mailbag just before it hits the ground is 14.57 m/s.
Explanation:
Given that,
Acceleration = 2 m/s
Time = 3 sec
We need to calculate the velocity of mailbag
Using equation of motion
[tex]v=u+at[/tex]
Put the value into the formula
[tex]v=0+2\times3[/tex]
[tex]v = 6 m/s[/tex]
We need to calculate the height at which the mailbag dropped
Using equation of motion
[tex]H=ut+\dfrac{1}{2}at^2[/tex]
Put the value into the formula
[tex]H=0+\dfrac{1}{2}\times2\times(3)^2[/tex]
[tex]H=9\ m[/tex]
We need to calculate the velocity of the mailbag just before it hits the ground
Using equation of motion
[tex]v^2= u^2+2gh[/tex]
[tex]v=\sqrt{u^2+2gh}[/tex]
Put the value into the formula
[tex]v=\sqrt{6^2+2\times9.8\times9}[/tex]
[tex]v=14.57\ m/s[/tex]
Hence, The velocity of the mailbag just before it hits the ground is 14.57 m/s.
