With only the drop height and the time we can use the kinematic equation:

We assume the initial velocity is  because the ball is dropped, plug in the height for , the time from drop to hitting the ground for , and then the only unknown variable is  which will be the experimental value Earth's gravitational acceleration.

We use the kinematic equation:

Plugging in  for ,  for  because it was dropped from rest, and  for  we have:

In the kinematic equation:

When the first term right of the equals sign goes to zero because the penny was dropped from rest, we see that the distance in a constant acceleration  is related to the time of acceleration  via:

because the fraction and acceleration are constant.

So when  is tripled from  to  (), the distance increases ninefold ()

Horizontal velocity does not affect gravity's force on an object so both the bullet and the ball are accelerated downward at the same gravitational acceleration and they will hit the ground at the same time.

So for this first we have to set up a relation between John (1) and Dave's (2) position. Keeping in mind that Dave is traveling the opposite distance of John.

Where x is the total distance between the two points. Then we solve it for time and plug it in to John's speed to find his distance after the elapsed time.

where is John's total distance traveled. Then we plug in our values

The free body diagram of the system is shown below:

 is the normal force on the block, and  is the weight of the block.

Since  is a component of , we can represent it as:

If you're confused as why it's cosine and not sine, think about the system practically. The flatter the slope is, the greater the normal force. The smaller an angle becomes (creating a flatter slope), the greater the value of cosine becomes, and subsequently the greater the normal force becomes.

Now we can simply plug in our given values:

First, calculate the gravitational force acting on the rock.

The exerts a force of downward, meaning that if the person exerted at least , then he or she would have been able to lift it up. Instead, the person applied only . This means that the person needed to apply  of additional force to lift the rock.

There are two forces in play for this scenario: the first is gravity and the second is friction. We can use Newton's second law to solve this problem:

Substituing in the two forces we just mentioned:

Note that the force of friction is subtracted because it is in the opposite direction of the force of gravity. Now, substituting in expressions for our two forces, we get:

If you are unsure of whether to use cosine or sine for each force, think about the situation practically. The flatter the slope gets, the less the force of gravity will have an effect on moving the block down the plane, hence the use of the sine function. Also, the flatter the slope gets, the greater the normal force will become, hence the use of the cosine function.

Canceling out mass from the equation and rearranging to solve for the coefficient of friction, we get:

There are two forces in play in this scenario. The first is gravity and the second is friction. Both depend on the angle of the slope. Since the block is traveling at a constant rate, we know the the gravitational and frictional force in the direction of the slope cancel each other out, and the net force is zero (there is no acceleration). Therefore, we can write:

Substituting in expressions for each force, we get:

If you are unsure of whether to use cosine or sine for each force, think about the situation practically. The flatter the slope gets, the less the force of gravity will have an effect on moving the block down the plane, hence the use of the sine function. Also, the flatter the slope gets, the greater the normal force will become, hence the use of the cosine function.

Canceling out mass and solving for the angle on one side of the equation, we get:

This is an important property to know! When an object travels down a slope at a constant rate, the tangent of the angle of the slope is equal to the coefficient of kinetic friction.