Answers:
a) 9.035 s
b) -88.543 m/s
Explanation:
The described situation is related to vertical motion (especifically free fall) and the equations that will be useful are:
[tex]y=y_{o}+V_{o}t+\frac{1}{2}gt^{2}[/tex] (1)
[tex]V=V_{o}+gt[/tex] (2)
Where:
[tex]y=0[/tex] is the final height of the steel ball
[tex]y_{o}=400 m[/tex] is the initial height of the steel ball
[tex]V_{o}=0[/tex] is the initial velocity of the steel ball (it was dropped)
[tex]V[/tex] is the final velocity of the steel ball
[tex]t[/tex] is the time it takes to the steel ball to reach the ground
[tex]g=-9.8 m/s^{2}[/tex] is the acceleration due to gravity
Knowing this, let's begin with the answers:
We will use equation (1) with the conditions listed above:
[tex]0=y_{o}+\frac{1}{2}gt^{2}[/tex] (3)
Isolating [tex]t[/tex]:
[tex]t=\sqrt{\frac{-2y_{o}}{g}}[/tex] (4)
[tex]t=\sqrt{\frac{-2(400 m)}{-9.8 m/s^{2}}}[/tex] (5)
[tex]t=9.035 s[/tex] (6)
We will use equation (2) with the conditions explained above and the calculaated time:
[tex]V=gt[/tex] (7)
[tex]V=(-9.8 m/s^{2})(9.035 s)[/tex] (8)
[tex]V=-88.543 m/s[/tex] (9) The negative sign indicates the direction of the velocity is downwards
