Respuesta :
Explanation:
It is given that,
Initial height of the ball, [tex]x_o=17\ m[/tex]
As it stops, its position, x = 0
It is in contact with the ground for 19.0 ms before stopping, [tex]t=19\times 10^{-3}\ s[/tex]
Initial speed of the ball, u = 0
(a) Let a is the acceleration of the ball during the time it is in contact with the ground. Its can be calculated suing equation of kinematics as :
[tex]v^2-u^2=2a(x-x_o)[/tex]
a =-g
[tex]v^2=-2g(x-x_o)[/tex]
[tex]v^2=-2g(-17)[/tex]
[tex]v^2=-2\times 9.8\times (-17)[/tex]
v = 18.25 m/s
(b) The average acceleration can be calculated as :
[tex]a_{avg}=\dfrac{v-u}{t}[/tex]
[tex]a_{avg}=\dfrac{18.25-0}{19\times 10^{-3}}[/tex]
[tex]a_{avg}=960.52\ m/s^2[/tex]
As the value of acceleration is positive, so the ball is accelerating in upward direction.
