Answer:
K = 0.076 J
Explanation:
The height of the target, h = 0.860 Â m
The mass of the steel ball, m = 0.0120 kg
Distance moved, d = 1.50 m
We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.
[tex]h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}[/tex]
Put all the values,
[tex]t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s[/tex]
The velocity of the ball is :
[tex]v=\dfrac{1.5}{0.418}\\\\= $3.58\ m/s[/tex]
The kinetic energy of the ball is :
[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J[/tex]
So, the required kinetic energy is 0.076 J.
