<h3>
Answer:</h3>
7.53 m
<h3>
Explanation:</h3>
<u>We are given:</u>
Initial Horizontal Velocity of the Ball = 4.6 m/s
Initial Vertical Velocity of the Ball = 0 m/s
Height from which ball is kicked = 13.4 m
<u>Time taken by the ball to reach the ground:</u>
The ball has an initial vertical velocity of 0 m/s
it also has a downward acceleration of 10 m/s² due to gravity
<u>Solving for the time taken:</u>
s = ut + 1/2(at²) [second equation of motion]
replacing the values
13.4 = (0)(t) + 1/2 (10)(t²)
13.4 = 5t²
t² = 13.4/5 [dividing both sides by 5]
t² = 2.68
t = 1.637 seconds [taking the square root of both sides]
<u>Horizontal distance covered by the ball:</u>
Since there are no horizontal opposing forces on the ball,
the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground
We calculated that the ball will hit the ground in 1.637 seconds
<u>Distance covered:</u>
s = ut + 1/2 (at²) [seconds equation of motion]
s = ut [since a = 0m/s² in the horizontal plane]
replacing the values
s = 4.6 * 1.637
s = 7.53 m
Hence, the ball landed 7.53 m from the cliff
