The initial velocity of the ball is 1.01 m/s
Explanation:
The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:
- A uniform horizontal motion with constant horizontal velocity
- A vertical accelerated motion with constant acceleration ([tex]g=9.8 m/s^2[/tex], acceleration due to gravity)
We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation
[tex]s=ut+\frac{1}{2}gt^2[/tex]
where
s = 1.20 m is the vertical displacement (the height of the desk)
u = 0 is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex]
t is the time of flight
Solving for t,
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.20)}{9.8}}=0.495 s[/tex]
Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of
d = 0.50 m
in a time
t = 0.495 s
Therefore, since the horizontal velocity is constant, we can calculate it as
[tex]v_x = \frac{d}{t}=\frac{0.50}{0.495}=1.01 m/s[/tex]
So, the ball rolls off the table at 1.01 m/s.
Learn more about projectile motion:
brainly.com/question/8751410
#LearnwithBrainly
