A block starts at rest and slides down a fric-

tionless track. It leaves the track horizontally,
flies through the air, and subsequently strikes
the ground.

What horizontal distance does the block
travel in the air?
Answer in units of m.

What is the speed of the block when it hits
the ground?
Answer in units of m/s.

To solve these questions, we can utilize the principle of conservation of energy. At the highest point of the block's path, it has maximum potential energy (PE) and no kinetic energy (KE). At the instant before it hits the ground, it has maximum KE and no PE. Since there is no friction, there is no loss of mechanical energy during the motion.

1) Horizontal Distance traveled in the air:
We can find this distance by analyzing the vertical motion of the block. Assuming the height of the starting point is h, and the height of the impact point on the ground is 0, we can use the conservation of energy principle to relate the initial potential energy (PE) to the final kinetic energy (KE):

PE = KE

mgh = (1/2)mv^2

Here, m is the mass of the block, g is the acceleration due to gravity, and v is the speed of the block at the moment it hits the ground. We can cancel out the mass m from both sides of the equation:

gh = (1/2)v^2

Solving for h:

h = (1/2)(v^2)/g

Since the horizontal distance traveled is the same as the horizontal distance covered in free fall, we can use the equation of motion for vertical motion:

h = (1/2)gt^2

Solving for t:

t = sqrt(2h/g)

The horizontal distance traveled in the air is given by:

d = vt,

where v is the horizontal velocity of the block. Since there is no horizontal force acting on the block, the horizontal velocity remains constant throughout its flight. Let's assume this velocity as v_horiz.

Therefore, substituting the value of t:

d = v_horiz * sqrt(2h/g)

2) Speed of the block when it hits the ground:
Since there is no loss of mechanical energy, the final kinetic energy of the block is equal to the initial potential energy at the top of the track:

PE_top = KE_bottom

mgh = (1/2)mv^2

Here, h is the height of the starting point, g is the acceleration due to gravity, and v is the speed of the block at the moment it hits the ground. We can cancel out the mass m from both sides of the equation:

gh = (1/2)v^2

Solving for v:

v = sqrt(2gh)

Now, you can substitute the values of g and h into the equation to find the horizontal distance traveled in the air and the speed of the block when it hits the ground.

To determine the horizontal distance the block travels in the air, we need to consider the motion of the block as it leaves the track and flies through the air.

The key concept to apply here is the conservation of horizontal velocity. Since the track is frictionless, there are no horizontal forces acting on the block once it leaves the track. Therefore, the horizontal component of the block's velocity remains constant during its flight.

To find the horizontal distance traveled by the block, we need to first determine the time it spends in the air. We can use the vertical motion of the block to find the time of flight.

Let's assume that the block falls from a height 'h'. We can calculate the time it takes for the block to strike the ground using the kinematic equation:

h = (1/2) * g * t^2

Where:
h = height from which the block falls (initial height)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken to reach the ground

Rearranging the equation to solve for 't', we get:

t = sqrt(2h / g)

Once we have the time of flight ('t'), we can use it to find the horizontal distance traveled by the block. Since the horizontal component of the block's velocity remains constant, we can use the equation:

horizontal distance = horizontal velocity * time

However, we need to determine the horizontal velocity of the block when it leaves the track. As the track is frictionless, the block only has a vertical component of velocity due to the force of gravity. The horizontal velocity remains unchanged.

To find the horizontal velocity, we need to consider the vertical component of the block's velocity when it leaves the track. Assuming the block starts from rest at the top of the track, the vertical velocity at that point is zero. This means the block falls freely under the influence of gravity.

Using the equation for vertical motion, we can find the vertical component of velocity when the block reaches the height 'h':

vf = sqrt(2gh)

Where:
vf = final vertical velocity (at height 'h')
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height from which the block falls

Since there are no horizontal forces acting, the horizontal component of velocity remains unchanged. Therefore, the horizontal velocity when the block leaves the track is the same as the horizontal velocity just before it reaches the height 'h'.

Finally, we can calculate the horizontal distance traveled by the block by multiplying the horizontal velocity by the time of flight:

horizontal distance = horizontal velocity * time

As for the speed of the block when it hits the ground, we can find it by calculating the total velocity just before impact. This velocity includes both the horizontal and vertical components.

The total velocity at the height 'h' can be found using the equation:

vt = sqrt(v_horizontal^2 + v_vertical^2)

Where:
vt = total velocity
v_horizontal = horizontal velocity
v_vertical = vertical velocity at height 'h'

Once we have the total velocity at the height 'h', we can use it to find the speed of the block when it hits the ground. Since the block only has a vertical velocity in the upward direction just before impact, the speed is equal to the magnitude of the vertical velocity:

speed = |v_vertical|

Now that we understand the principles and equations involved, we can proceed to plug in the given values and calculate the answers.

You have to figure the time in the air from the height of the drop from the end of the track to the ground.

Given the time, you know the horizontal distance (speed*time).

I will be happy to critique your thinking, or work.