A stone is thrown 60 degrees below the horizontal at 22 ft/sec and 20 meters above the ground. Using the energy equations, determine its velocity when it hits the ground.
Are you sure the stone was thrown at an angle below the horizontal? If so,
h(t) = 20 - (22*sin60)t - 4.9t^2
solve for t when h=0.
the vertical speed is -22*sin60 - 9.8t
the horizontal speed is a constant 22*cos60
so, knowing the final components of the velocity, you can figure its magnitude and direction.
ft/sec?
If it is you need to convert as well.
To determine the velocity of the stone when it hits the ground, we can use the principle of conservation of energy.
The initial potential energy of the stone (PEi) is given by the formula:
PEi = m * g * h
where m is the mass of the stone, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the initial height of the stone (20 meters).
The initial kinetic energy of the stone (KEi) is given by the formula:
KEi = 0.5 * m * v^2
where v is the initial velocity of the stone (22 ft/s).
Since we need to work with the same unit system, let's convert the initial velocity from ft/s to m/s:
22 ft/s = 6.71 m/s (approximately)
Now, we can equate the initial potential energy and initial kinetic energy to the final kinetic energy (KEf) when the stone hits the ground, and solve for the final velocity (vf).
PEi + KEi = KEf
m * g * h + 0.5 * m * v^2 = 0.5 * m * vf^2
Notice that mass (m) appears on both sides of the equation, so we can cancel it out:
g * h + 0.5 * v^2 = 0.5 * vf^2
Now we can solve for vf:
vf^2 = 2 * (g * h + 0.5 * v^2)
vf = sqrt(2 * (g * h + 0.5 * v^2))
Substituting the given values:
vf = sqrt(2 * (9.8 m/s^2 * 20 m + 0.5 * 6.71 m/s^2 * 6.71 m/s^2))
Simplifying:
vf = sqrt(2 * (196 m^2/s^2 + 22.5 m^2/s^2))
vf = sqrt(2 * 218.5 m^2/s^2)
vf = sqrt(437 m^2/s^2)
vf ≈ 20.9 m/s
Therefore, the velocity of the stone when it hits the ground is approximately 20.9 m/s.