Answer:
The ball will be at 700 m above the ground.
Explanation:
We can use the following kinematic equation
[tex]y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2[/tex].
where y(t) represent the height from the ground. For our problem, the initial height will be:
[tex]y_0 \ = \ 1000 m[/tex].
The initial velocity:
[tex]v_0 = - 20 \frac{m}{s}[/tex],
take into consideration the minus sign, that appears cause the ball its thrown down. The same minus appears for the acceleration:
[tex]a=-10\frac{m}{s}[/tex]
So, the equation for our problem its:
[tex]y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2[/tex].
Taking t=6 s:
[tex]y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2[/tex].
[tex]y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2[/tex].
[tex]y(6 \ s) = \ 1000 m \ - 120 m - 180 m[/tex].
[tex]y(6 \ s) = \ 1000 m \ - 300 m[/tex].
[tex]y(6 \ s) = \ 700 m [/tex].
So this its the height of the ball 6 seconds after being thrown.
