Answer:
After 1.00 sec the speed of the ball will be [tex]18.4 \frac{m}{s}[/tex]
Explanation:
The equation to find the speed at a given time in a vertical thrown is:
[tex]s=s_0-g(t-t_0)[/tex]
Where [tex]s_0[/tex] is the speed at a time [tex]t_0[/tex], and [tex]g[/tex] is the accelaration due to gravity.
Then we can calculate the speed of the ball after 1.00 sec replacing the given data:
[tex]s=s_0-g(t-t_0)[/tex]
[tex]s_1=28.2 \frac{m}{s} -9.8 \frac{m}{s^2} (1 s-0s)[/tex]
[tex]s_1=18.4 \frac{m}{s}[/tex]
Then, after 1.00 sec the speed of the ball will be [tex]18.4 \frac{m}{s}[/tex]
The used equation can be found integrating the definition of acceleration as the derivative of speed with respect of time. Like this:
[tex]a=\frac{ds}{dt}[/tex]
[tex]adt=ds[/tex]
Integrating both sides:
[tex]\int\limits^t_{t_0} {a} \, dt = \int\limits^s_{s_0} {ds}[/tex]
[tex](at) \Big|_{t_0}^t = (s)\Big|_{s_0}^s[/tex]
[tex]a(t-t_0)=(s-s_0)[/tex]
[tex]s=s_0+a(t-t_0)[/tex]
In te case of a vertical thrown, [tex]a=-g[/tex] so
[tex]s=s_0-g(t-t_0)[/tex]
