The ball's initial velocity must be 28.0 m/s
Explanation:
The motion of the ball in this problem is a projectile motion, so it follows a parabolic path, consisting of two separate motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
First, we study the vertical motion of the ball: since it is a uniformly accelerated motion, we can use the suvat equation to find the time it takes for the ball to reach the ground,
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = 40 m is the vertical displacement, the height of the cliff (we chose downward as positive direction)
u = 0 is the initial vertical velocity of the ball
t is the time of flight of the ball
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t, we find:
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(40)}{9.8}}=2.86 s[/tex]
Now we can analyze the horizontal motion: since this is a uniform motion, the horizontal speed is constant, and it is given by
[tex]v_x = \frac{d}{t}[/tex]
where:
d = 80 m is the horizontal distance covered by the ball
t = 2.86 s is the time of flight
Substituting,
[tex]v_x = \frac{80}{2.86}=28.0 m/s[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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