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