Answer:
32.3 m/s
Explanation:
The ball follows a projectile motion, where:
- The horizontal motion is a uniform motion at costant speed
- The vertical motion is a free fall motion (constant acceleration)
We start by analyzing the horizontal motion. The ball travels horizontally at constant speed of
[tex]v_x = 50 m/s[/tex]
and it covers a distance of
d = 165 m
So, the total time of flight of the ball is
[tex]t=\frac{d}{v_x}=\frac{165}{50}=3.3 s[/tex]
In order to find the vertical velocity of the ball, we have now to analyze its vertical motion.
The vertical motion is a free-fall motion, so the ball is falling at constant acceleration; therefore we can use the following suvat equation:
[tex]v_y = u_y +at[/tex]
where
[tex]v_y[/tex] is the vertical velocity at time t
[tex]u_y=0[/tex] is the initial vertical velocity
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity (taking downward as positive direction)
Substituting t = 3.3 s (the time of flight), we find the final vertical velocity of the ball:
[tex]v=0 + (9.8)(3.3)=32.3 m/s[/tex]
