A small hockey puck slides without friction over the icy hill as shown above, and lands 14.9 m from the foot of the cliff with no air resistance.What was its speed v0 at the bottom of the hill?Express your answer in m/s, to at least one digit after the decimal point.
Question
A small hockey puck slides without friction over the icy hill as shown above, and lands 14.9 m from the foot of the cliff with no air resistance.What was its speed v0 at the bottom of the hill?Express your answer in m/s, to at least one digit after the decimal point.
Solution
To find the speed v0 of the hockey puck at the bottom of the hill, we can use the principle of conservation of energy.
First, let's consider the initial potential energy of the puck at the top of the hill. This potential energy is given by the equation PE = mgh, where m is the mass of the puck, g is the acceleration due to gravity, and h is the height of the hill. However, since there is no information given about the height of the hill, we cannot directly calculate the potential energy.
Next, let's consider the final kinetic energy of the puck at the bottom of the hill. The kinetic energy is given by the equation KE = (1/2)mv^2, where m is the mass of the puck and v is its velocity.
Since there is no friction or air resistance, we can assume that the total mechanical energy of the puck is conserved. Therefore, the initial potential energy at the top of the hill is equal to the final kinetic energy at the bottom of the hill.
Setting the equations for potential energy and kinetic energy equal to each other, we have:
mgh = (1/2)mv^2
The mass of the puck cancels out, and we are left with:
gh = (1/2)v^2
Solving for v, we have:
v = sqrt(2gh)
Since we do not have the value of h, we cannot directly calculate the speed v0 at the bottom of the hill.
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