Use the following kinematic equation where the initial velocity is , final velocity is , and the distance traveled is .

We can use the values in the question to solve for the acceleration.

We rearrange the equation to solve for the acceleration.

The first step to solving this problem will be to find the velocity of the ball at the point when the picture is taken. We know the initial velocity of the ball (zero), the displacement, and the acceleration. Using the appropriate kinematics formula, we can solve for the final velocity.

Now that we know how fast the ball is traveling when the picture is taken, we can find the distance it travels while the shutter is open. This distance will become a motion blur, making the vertical diameter of the ball appear stretched.

During the exposure period, the ball will travel , or . The diameter of the ball in the vertical direction will appear to be distorted by this distance.

This is projectile motion in the vertical direction only, subject to the equation of motion: .

For this discussion, one can define the downward direction as negative. For projectile motion,  (gravitational acceleration, or ).

In this case, the ball ends up  below where is started, so .

The initial velocity, , is  (upward, thus positive).

With all this, the projectile motion equation becomes:

This can be solved for  using the quadratic formula:

 The result is:

 (or ).

Only the positive answer option is physically possible, and is thus our correct answer.

First we will find the time required for the ball to reach the ground. Since the ball is thrown horizontally, it has no initial vertical component. We use the following equation to solve for the total flight time:

We are given the change in height, initial velocity, and acceleration. Using these values, we can solve for the time. Note that the change in height will be negative, since the ball is traveling downward.

Finally, we use the horizontal velocity to find the distance traveled in . Remember that the horizontal velocity remains constant during projectile motion.

Since we know the change in velocity of the puck, we can determine the work done by friction by using the work-energy theorem:

The work in this question is done by friction, so we can write:

Substituting in expressions for friction and kinetic energy, we get:

The normal force is from gravity, so we can write:

Rearranging for the coefficient of kinetic friction we get:

The equation needed is

Speed is proportional to the absolute value of the slope. Boomer's speed has its largest value  during the intervals from  and from . Note that speed is independent of direction.

Boomer is at rest during periods when the slope of the position versus time graph is zero.

Boomer's position does not change during the interval when he is at rest. This is the final interval.

Boomer's speed is the ratio of the distance traveled to the time. It is  during both the first and and third intervals.