Answer:
(a) v = 5.42m/s
(b) vo = 4.64m/s
(c) a = 2874.28m/s^2
(d) Δy = 5.11*10^-3m
Explanation:
(a) The velocity of the ball before it hits the floor is given by:
[tex]v=\sqrt{2gh}[/tex] (1)
g: gravitational acceleration = 9.8m/s^2
h: height where the ball falls down = 1.50m
[tex]v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}[/tex]
The speed of the ball is 5.42m/s
(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.
You use the following formula:
[tex]h_{max}=\frac{v_o^2}{2g}[/tex] (2)
vo: velocity of the ball where it starts its motion upward
You solve for vo and replace the values of the parameters:
[tex]v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}[/tex]
The velocity of the ball is 4.64m/s
(c) The acceleration is given by:
[tex]a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}[/tex]
[tex]a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}[/tex]
The acceleration of the ball is 2874.28/s^2
(d) The compression of the ball is:
[tex]\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m[/tex]
THe compression of the ball when it strikes the floor is 5.11*10^-3m
