given two continuous time signals x(t) and h(t), if their convultion is denoted by y(t) , then the convultion of x(t-1) and h(t+
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The distance below the top of the cliff that the two balls cross paths is 7.53 meters.
<u>Given the following data:</u>
- Initial velocity = 0 m/s (since the ball is dropped from rest).
- Height = 30 meters.
<u>Scientific data:</u>
- Acceleration due to gravity (a) = 9.8
.
To determine how far (distance) below the top of the cliff that the two balls cross paths, we would apply the third equation of motion.
<h3>How to calculate the velocity.</h3>Mathematically, the third equation of motion is given by this formula:
<u>Where:</u>
- V is the final velocity.
- U is the initial velocity.
- a is the acceleration.
- S is the distance covered.
Substituting the parameters into the formula, we have;
V = 24.25 m/s.
<u>Note:</u> The final velocity of the first ball becomes the initial velocity of the second ball.
The time at which the two balls meet is calculated as:
Time = 1.24 seconds.
The position of the ball when it is dropped from the cliff is calculated as:
Lastly, the distance below the top of the cliff is calculated as:
Distance = 7.53 meters.
Read more on distance here: brainly.com/question/10545161
