A Zip-liner is gliding along a horizontal wire over a lake. They must release themselves from the wire and drop down to the water to get off the zip-line. They are 6.50 m above the water and release themselves when they are travelling at 2.20 m/s. How fast are they travelling when they hit the water?
Y^2 = Yo^2 + 2g*h = 0 + 19.6*6.5 = 127.4, Y = 11.29 m/s.
V = sqrt(Xo^2+Y^2) = sqrt(2.2^2 + 11.29^2) = 11.50 m/s
To determine the speed at which the zip-liner hits the water, classical physics principles can be applied. The conservation of mechanical energy is particularly relevant in this scenario.
1. Start by calculating the initial potential energy of the zip-liner when they release themselves. The potential energy is given by the equation PE = mgh, where m is the mass of the zip-liner, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the water. In this case, m and g are not provided, but we can assume a relatively constant value of g.
Given:
h = 6.50 m
g = 9.8 m/s^2
PE = mgh = m * 9.8 * 6.50
2. Next, find the initial kinetic energy of the zip-liner. The kinetic energy is given by the equation KE = (1/2)mv^2, where m is the mass of the zip-liner and v is the initial velocity at the time of release.
Given:
v = 2.20 m/s
KE = (1/2)mv^2 = (1/2) * m * (2.20)^2
3. As per the conservation of mechanical energy, the initial potential energy is equal to the final kinetic energy, neglecting any energy losses due to air resistance. Therefore, we can set the potential energy equal to the kinetic energy:
m * 9.8 * 6.50 = (1/2) * m * (2.20)^2
4. Solve the equation for the mass "m" by dividing both sides of the equation by (9.8 * 6.50):
9.8 * 6.50 = (1/2) * (2.20)^2 * m
m = (9.8 * 6.50) / [(1/2) * (2.20)^2]
5. Calculate the mass of the zip-liner.
m = 206.68 kg
6. To find the speed at which the zip-liner hits the water, we need to determine the velocity at that point. We can do this using the equation v^2 = u^2 + 2gh, where u is the initial velocity (provided), g is the acceleration due to gravity, and h is the height from which the zip-liner is released.
Given:
u = 2.20 m/s
h = 6.50 m
g = 9.8 m/s^2
v^2 = u^2 + 2gh
v^2 = (2.20)^2 + 2 * 9.8 * 6.50
v^2 = 4.84 + 127.4
v^2 = 132.24
Taking the square root of both sides:
v = √132.24
7. Calculate the final velocity.
v ≈ 11.51 m/s
Hence, the zip-liner will be traveling at approximately 11.51 m/s when they hit the water.