A puck slides across a smooth, level tabletop at height H at a constant speed Vo. It slides off the edge of the table and hits the floor a distance x away. What is the algebraic relationship between the distances x and H?
time in air:
h=4.9 t^2
t= sqrt(H/4.9)
distance traveled=vo*t
x=vo*sqrt(H/4.9)
To determine the algebraic relationship between the distances x and H, we can analyze the motion of the puck.
When the puck slides off the edge of the table, it becomes a projectile in free fall. In the absence of air resistance, the horizontal and vertical motions of the puck are independent of each other. This means we can consider them separately.
Since the horizontal speed Vo remains constant, the horizontal distance traveled x is solely determined by the time of flight. We can find the time of flight by using the equation:
x = Vo * t,
where t is the time of flight.
In the vertical direction, the puck is influenced by the force of gravity. The height H is related to the time of flight t by the equation:
H = (1/2) * g * t^2,
where g is the acceleration due to gravity.
Now, we can solve each equation for t and substitute it into the other equation to eliminate the variable t.
From the first equation, we get:
t = x / Vo.
Substituting this expression for t into the second equation, we have:
H = (1/2) * g * (x / Vo)^2.
Simplifying and rearranging, we find:
2H = (g / Vo^2) * x^2.
Therefore, the algebraic relationship between the distances x and H is:
x^2 = (2H * Vo^2) / g.
To determine the algebraic relationship between the distances x and H, we can start by analyzing the motion of the puck.
When the puck slides off the edge of the table, it will experience two types of motion: horizontal motion and vertical motion. The horizontal motion will not be influenced by the height of the table, but the vertical motion will be affected by the height H.
Let's consider the vertical motion first. The puck will fall freely under the influence of gravity, and we can use the equations of motion to describe its vertical displacement.
The equation for the vertical displacement (y) of an object in free fall is given by:
y = Vo*t - 0.5*g*t^2
Where:
Vo is the initial vertical velocity (0 since it is falling),
t is the time,
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the vertical displacement of the puck is equal to the height H, we can write the equation as:
H = -0.5*g*t^2
Now let's consider the horizontal motion. The puck slides off the table with a constant speed Vo, so there is no horizontal acceleration. The equation for horizontal distance traveled (x) can be expressed as:
x = Vo*t
By combining the two equations (H = -0.5*g*t^2 and x = Vo*t), we can eliminate the variable t.
From the equation x = Vo*t, we can isolate t and express it as:
t = x / Vo
Substituting this value of t into the equation H = -0.5*g*t^2, we get:
H = -0.5*g*(x / Vo)^2
Therefore, the algebraic relationship between the distances x and H is:
H = -0.5*g*(x / Vo)^2