Answer:
a) v₀ = 9.2 m/s
b) y₀ = 7.9 m
Explanation:
The position of the balls is given by the equation:
[tex]y =- \frac{1}{2} gt^2 + v_0 t + y_0[/tex]
where:
acceleration g = 9.8 m/s²
time t
initial velocity v₀
initial height y₀
a) lets divide (a) in two parts:
1.part: How long will it take the second ball to fall down?
[tex]v_0 = 0, y = 0\\0=- \frac{1}{2} gt^2 + y_0\\ t = \sqrt{\frac{2y_0}{g}}[/tex]
2. part: At time t from part1 + 1.15s, the first ball should land on the ground.
[tex]y = 0, y_0 = 19.6, t = \sqrt{\frac{2y_0}{g}} + 1.15\\ 0 =- \frac{1}{2} gt^2 + v_0t + y_0[/tex]
This leaves only one unknown: v₀
[tex]v_0 =\frac{1}{t}(\frac{1}{2} gt^2 - y_0)\\ v_0 = 9.2 \frac{m}{s}[/tex]
b)again, lets divide in two parts
1.part: Where will ball1 be relative to ball2 in 1.15s:
[tex]t = 1.15s, v_0 = 8.6 m/s\\y= -\frac{1}{2} gt^2 + v_0t + y_0\\ \delta y = y - y_0 =v_0t -\frac{1}{2} gt^2[/tex]
and how fast will it go:
[tex]v' = -gt + v_0[/tex]
2.part: Now we can plug in to the equation for the position of the two balls. Let's start with the second ball first:
[tex]0 = -\frac{1}{2} gt^2 + y_0\\ y_0 = \frac{1}{2} gt^2[/tex]
Now let's use this result in the equation for the first ball:
[tex]0 = - \frac{1}{2} gt^2 + v't + y_0 + \delta y = - \frac{1}{2} gt^2 + v't + \frac{1}{2} gt^2 + \delta y\\ 0 = v't + \delta y\\ t =- \frac{\delta y}{v'} \\ y_0 = \frac{1}{2} g(\frac{\delta y}{v'})^2\\ y_0 = 7.9m[/tex]
