Answer:
0.784 m
Explanation:
The motion of the ball is a projectile motion, which consists of two independent motions:
- a horizontal uniform motion, with constant speed of [tex]v_0 = 1.6 m/s[/tex]
- a vertical accelerated motion, with constant acceleration [tex]g=-9.8 m/s^2[/tex] towards the ground and initial position [tex]y_0 = 1.2 m[/tex]
By looking at the vertical motion, we can find the time t at which the ball hits the ground, which is the time t at which the vertical position y becomes zero:
[tex]y(t)=0\\y(t)=y_0 +\frac{1}{2}gt^2\\0=y_0 +\frac{1}{2}gt^2\\t=\sqrt{-\frac{2 y_0}{g}}=\sqrt{-\frac{2\cdot 1.2 m}{-9.8 m/s^2}}=0.49 s[/tex]
And now we can use the horizontal motion to find the total horizontal distance traveled during this time interval:
[tex]S_x = v t=(1.6 m/s)(0.49 s)=0.784 m[/tex]
