The horizontal distance covered by the ball is 75.8 m
Explanation:
The motion of the ball in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
First of all, we consider the vertical motion to find the time of flight of the ball. Using the suvat equation:
[tex]s=ut+\frac{1}{2}gt^2[/tex]
where
s = 45 m is the vertical displacement of the ball
t is the time of flight
u = 0 is the initial vertical velocity of the ball
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t,
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s[/tex]
Now we know that the ball moves horizontally with a constant velocity of
[tex]v_x = 25 m/s[/tex]
So, the horizontal distance covered by the ball during its flight is
[tex]d=v_x t = (25)(3.03)=75.8 m[/tex]
So the ball lands 75.8 m far from the base of the cliff.
Learn more about projectile motion:
brainly.com/question/8751410
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