Using energy considerations and assuming negligible air resistance, find the speed of a rock just before it hits the water if it is thrown from a bridge 20.2 m above water with an initial speed of 13.4 m/s? (Verify that the speed is independent of the direction the rock is thrown.)
1/2 mvi^2 + mgh = 1/2mvf^2
m cancels
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To find the speed of the rock just before it hits the water, we can use the principle of conservation of mechanical energy. The total mechanical energy of the rock at the top of the bridge will be equal to the total mechanical energy of the rock just before it hits the water.
Let's assume that the reference level for gravitational potential energy is at the water level. At the top of the bridge, the rock has both potential energy and kinetic energy:
Initial potential energy = mgh, where m is the mass of the rock, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the bridge (20.2 m).
Initial kinetic energy = 1/2 mv^2, where v is the initial speed of the rock (13.4 m/s).
Therefore, the total mechanical energy at the top of the bridge is:
Total initial energy = mgh + 1/2 mv^2
At the water level, the rock only has kinetic energy, as its potential energy is zero due to being at the reference level. The total mechanical energy just before hitting the water is:
Total final energy = 1/2 mv²_final
Since energy is conserved, we have:
Total initial energy = Total final energy
mgh + 1/2 mv^2 = 1/2 mv²_final
Canceling out m and rearranging the equation, we get:
gh + 1/2 v^2 = 1/2 v²_final
Simplifying further:
gh = 1/2 v²_final - 1/2 v^2
v²_final = 2gh + v^2
Finally, we can solve for the final speed v_final by taking the square root of both sides:
v_final = √(2gh + v^2)
Plugging in the given values:
v_final = √(2 * 9.8 m/s^2 * 20.2 m + (13.4 m/s)^2)
v_final ≈ √(396.04 + 179.56)
v_final ≈ √575.6
v_final ≈ 23.99 m/s
Therefore, the speed of the rock just before it hits the water is approximately 23.99 m/s. This speed is independent of the direction the rock is thrown.
To find the speed of the rock just before it hits the water, we can use the principle of conservation of mechanical energy. The mechanical energy of the rock is conserved, assuming that there is no air resistance and neglecting other non-conservative forces.
At the top of the bridge, the rock has potential energy due to its height above the water. At the bottom, just before it hits the water, the rock has both potential and kinetic energy.
The potential energy at the top is given by the formula:
PE = m * g * h
Where:
m = mass of the rock
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height above water (20.2 m)
So, the potential energy at the top is:
PE_top = m * g * h
The kinetic energy just before the rock hits the water is given by the formula:
KE = 0.5 * m * v^2
Where:
v = speed of the rock just before hitting the water
From the conservation of mechanical energy principle, we know that the total mechanical energy is conserved throughout the motion. Therefore, the potential energy at the top is equal to the kinetic energy just before hitting the water:
PE_top = KE
Substituting the equations for potential energy and kinetic energy:
m * g * h = 0.5 * m * v^2
The mass of the rock (m) cancels out, and we can solve for v:
g * h = 0.5 * v^2
v^2 = 2 * g * h
v = √(2 * g * h)
Now, we can calculate the speed of the rock just before hitting the water by substituting the values of g and h:
v = √(2 * 9.8 * 20.2) = 19.9 m/s
Therefore, the speed of the rock just before hitting the water is approximately 19.9 m/s.
To verify that the speed is independent of the direction the rock is thrown, we can consider that the speed depends only on the height (h) and acceleration due to gravity (g). These values remain the same irrespective of the direction of the throw. Hence, the speed of the rock just before hitting the water is the same regardless of whether it is thrown upwards or downwards.