Respuesta :
Explanation:
Given that,
Mass of the ball, m = 1.2 kg
Initial speed of the ball, u = 10 m/s
Height of the floor from ground, h = 32 m
(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2\times 9.8\times 32}[/tex]
v = -25.04 m/s (negative as it rebounds)
The impulse acting on the ball is equal to the change in momentum. It can be calculated as :
[tex]J=m(v-u)[/tex]
[tex]J=1.2\times (-25.04-10)[/tex]
J = -42.048 kg-m/s
(b) Time of contact, t = 0.02 s
Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :
[tex]J=\dfrac{F}{t}[/tex]
[tex]F=J\times t[/tex]
[tex]F=42.048\times 0.02[/tex]
F = 0.8409 N
Hence, this is the required solution.
