Answer:
7.9 m/s
Explanation:
When both balls collide, they have spent the same time for their motions.
Motion of steel ball
This is purely under gravity. It is vertical.
Initial velocity, <em>u </em>= 0 m/s
Distance, <em>s</em> = 4.0 m - 1.2 m = 2.8 m
Acceleration, <em>a</em> = g
Using the equation of motion
![s = ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%20%3D%20ut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
![2.8 \text{ m} = 0+\dfrac{gt^2}{2}](https://tex.z-dn.net/?f=2.8%20%5Ctext%7B%20m%7D%20%3D%200%2B%5Cdfrac%7Bgt%5E2%7D%7B2%7D)
![t = \sqrt{\dfrac{5.6}{g}}](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B%5Cdfrac%7B5.6%7D%7Bg%7D%7D)
Motion of plastic ball
This has two components: a vertical and a horizontal.
The vertical motion is under gravity.
Considering the vertical motion,
Initial velocity, <em>u </em>= ?
Distance, <em>s</em> = 1.2 m
Acceleration, <em>a</em> = -<em>g </em> (It is going up)
Using the equation of motion
![s = ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%20%3D%20ut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
![1.2\text{ m} = ut-\frac{1}{2}gt^2](https://tex.z-dn.net/?f=1.2%5Ctext%7B%20m%7D%20%3D%20ut-%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
Substituting the value of <em>t</em> from the previous equation,
![1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}](https://tex.z-dn.net/?f=1.2%5Ctext%7B%20m%7D%20%3D%20u%5Csqrt%7B%5Cdfrac%7B5.6%7D%7Bg%7D%7D-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20g%5Ctimes%5Cdfrac%7B5.6%7D%7Bg%7D)
![u\sqrt{\dfrac{5.6}{g}} = 4.0](https://tex.z-dn.net/?f=u%5Csqrt%7B%5Cdfrac%7B5.6%7D%7Bg%7D%7D%20%3D%204.0)
Taking <em>g</em> = 9.8 m/s²,
![u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}](https://tex.z-dn.net/?f=u%20%3D%20%5Cdfrac%7B4.0%7D%7B0.756%7D%20%3D%205.29%20%5Ctext%7B%20m%2Fs%7D)
This is the vertical component of the initial velocity
Considering the horizontal motion which is not accelerated,
horizontal component of the initial velocity is horizontal distance ÷ time.
![u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}](https://tex.z-dn.net/?f=u_h%20%3D%20%5Cdfrac%7B4.4%5Ctext%7B%20m%7D%7D%7B0.756%5Ctext%7B%20s%7D%7D%20%3D%205.82%5Ctext%7B%20m%2Fs%7D)
The initial velocity is
![v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}](https://tex.z-dn.net/?f=v_i%20%3D%20%5Csqrt%7Bu%5E2%2Bu_h%5E2%7D%20%3D%20%5Csqrt%7B%285.29%5Ctext%7B%20m%2Fs%7D%29%5E2%2B%285.82%5Ctext%7B%20m%2Fs%7D%29%5E2%7D%20%3D%207.9%20%5Ctext%7B%20m%2Fs%7D)