Question

A ball is dropped from rest from the top of a cliff that is 30 m high. From ground

Answer 1

The distance below the top of the cliff that the two balls cross paths is 7.53 meters.

Given the following data:

• Initial velocity = 0 m/s (since the ball is dropped from rest).

• Height = 30 meters.

Scientific data:

• Acceleration due to gravity (a) = 9.8 $m/s^2$.

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.

How to calculate the velocity.

Mathematically, the third equation of motion is given by this formula:

$V^2 = U^2 +2aS$

Where:

• 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;

$V^2 = 0^2 +2(9.8) \times 30\\\\V^2 = 588\\\\V=\sqrt{588}$

V = 24.25 m/s.

Note: 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:

$Time = \frac{S}{U} \\\\Time = \frac{30}{24.25}$

Time = 1.24 seconds.

The position of the ball when it is dropped from the cliff is calculated as:

$y_1 = h-\frac{1}{2} at^2\\\\y_1 = 30-\frac{1}{2} \times 9.8 \times 1.24^2\\\\y_1 = 30-7.53\\\\y_1=22.47\;meters$

Lastly, the distance below the top of the cliff is calculated as:

$Distance = 30-22.47$

Distance = 7.53 meters.

Read more on distance here: brainly.com/question/10545161