Answer:
1.65 m
Explanation:
The motion of the ball is a projectile motion, so it is a parabolic motion with two independent motions:
- on the x-axis, a uniform motion with constant speed [tex]v_x=12.0 m/s[/tex]
- on the y-axis, a uniformly accelerated motion with constant acceleration [tex]a=9.8 m/s^2[/tex] downward.
First of all, we need to find the time t at which the ball has travelled halfway to the chatcher, i.e. at a distance of [tex]x=14.0/2=7.0 m[/tex]. This can be found by using the relationship between distance, time and velocity along the horizontal direction:
[tex]t=\frac{x}{v_x}=\frac{7.0 m}{12.0 m/s}=0.58 s[/tex]
So now we can move to the vertical motion, and we can calculate the distance covered vertically by the ball while falling for t=0.58 s, by using:
[tex]S=\frac{1}{2}at^2=\frac{1}{2}(9.8 m/s^2)(0.58 s)^2=1.65 m[/tex]
