Answer:
v₀ = 2.68 m/s.
[tex]v_{B}[/tex] = 3.33 m/s
d = 1.61 m
Explanation:
Searching on the internet I found the picture of the answer in which the height between point A and point B is equal to 0.2 m.
To determine the value of v₀ we need to use the following equation:
[tex] v_{f} = v_{0} - gt [/tex]
Taking the initial vertical velocity ([tex]v_{0_{y}}[/tex]) as [tex]v_{A_{y}}[/tex] and the final vertical velocity ([tex]v_{f_{y}}[/tex]) as [tex]v_{B_{y}}[/tex] we have:
[tex]v_{B_{y}} = v_{A_{y}} - gt[/tex]
We can express the time (t) in terms of the velocities of A and B:
[tex] t = \frac{v_{A_{y}} - v_{B_{y}}}{g} [/tex] (1)
Now, we can use the equation:
[tex]y_{f} = y_{0} + v_{A_{y}}t - \frac{1}{2}gt^{2}[/tex] (2)
By entering equation (1) into equation (2):
[tex]y_{f} = y_{0} + v_{A_{y}}*(\frac{v_{A_{y}} - v_{B_{y}}}{g}) - \frac{1}{2}g(\frac{v_{A_{y}} - v_{B_{y}}}{g})^{2}[/tex] (3)
Since [tex]v_{A_{x}} = v_{B_{x}} [/tex] because there is no acceleration in the horizontal movement, we have:
[tex] v_{A}sin(15) = v_{B}sin(12) [/tex]
[tex] v_{A} = \frac{v_{B}sin(12)}{sin(15)} [/tex] (4)
Taking:
[tex]y_{f}[/tex] = 0
[tex]y_{0}[/tex] = 0.2 m
g = 9.81 m/s²
[tex]v_{A_{y}} = v_{A}cos(15)[/tex]
[tex]v_{B_{y}} = v_{B}cos(12)[/tex]
Entering the above values and equation (4) into equation (3) we have:
[tex] -0.4 m*9.81 m/s^{2} = (\frac{v_{B}sin(12)cos(15)}{sin(15)})^{2} - (v_{B}cos(12))^{2} [/tex]
By solving the above quadratic equation:
[tex]v_{B} = 3.33 m/s [/tex]
Hence, the velocity of the ball [tex]v_{B}[/tex] is 3.33 m/s.
Now, we can find [tex]v_{A}[/tex] by using equation (4):
[tex]v_{A} = \frac{v_{B}sin(12)}{sin(15)} = \frac{3.33 m/s*sin(12)}{sin(15)} = 2.68 m/s [/tex]
Then, the value of v₀ of the ball is 2.68 m/s.
Finally, to find the distance between point A and point B we need to calculate the time by using equation (1):
[tex]t = \frac{v_{A_{y}} - (-v_{B_{y}})}{g} = \frac{v_{A_{y}} + v_{B_{y}}}{g}[/tex]
The minus sign in [tex]v_{B_{y}}[/tex] is because the vertical component of the vector is negative.
[tex] t = \frac{v_{A_{y}} + v_{B_{y}}}{g} = \frac{2.68 m/s*cos(15) + 3.33 m/s*cos(12)}{9.81 m/s^{2}} = 0.60 s [/tex]
Now, the distance is:
[tex] x = v_{0}*t = 2.68 m/s*0.60 s = 1.61 m [/tex]
I hope it helps you!
