The ball was kicked with a speed of 38.4 m/s.
Using s = ut + 1/2gt², we find the time it takes the ball to reach the ground. Where
- s = height of cliff = 75 m
- u = <u>initial</u> vertical velocity of ball = 0 m/s(since it is kicked horizontally)
- g = acceleration due to gravity = 9.8 m/s²
- t = time taken for ball to reach the ground.
Substituting the values of the variables into the equation, we have
s = ut + 1/2gt²
75 m = (0 m/s)t + 1/2 × 9.8 m/s² × t²
75 m = 0 m + (4.9 m/s²) t²
75 m = (4.9 m/s²)t²
Dividing both sides by 4.9, we have
t² = 75 m/4.9 m/s²
t² = 15.306 s²
Taking square-root of both sides, we have
t = √15.306 s²
t = 3.91 s
The horizontal distance covered by the ball d = vt where
- v = <u>horizontal</u> speed of ball and,
- t = time taken for the ball to reach the ground.
Since we require v, we make it subject of the formula.
So, v = d/t
Given that d = 150 m and t = 3.91 s, substituting these into v, we have
v = d/t
v = 150 m/3.91 s
v = 38.4 m/s
So, the ball was kicked with a speed of 38.4 m/s.
Learn more about motion under gravity here:
brainly.com/question/25271752
