Answer:
0.910 m
Explanation:
The momentum of the ball before it hits the ground is its final momentum.
[tex]p_f = mv_f[/tex]
m is the mass of the ball and [tex]v_f[/tex] is the final velocity.
[tex]v_f = \dfrac{p_f}{m} = \dfrac{0.740\text{ kg m/s}}{0.175\text{ kg}} = 4.223\text{ m/s}[/tex]
The motion of the ball is under gravity, dropping from rest.
The initial velocity, [tex]v_i[/tex], is 0 m/s because it is dropped.
Acceleration of gravity, g, is 9.8 m/s².
The height is denoted by h.
We use the equation of motion:
[tex]v_f^2 = v_i^2 + 2gh[/tex]
(We are taking g as positive here).
[tex]h= \dfrac{v_f^2 - v_i^2}{2g}[/tex]
[tex]h= \dfrac{(4.223^2 - 0^2)\text{ m}^2\text{/s}^2}{2\times 9.8\text{ m/s}^2} = 0.910 \text{ m}[/tex]
