A boy of mass m takes a running start and jumps on his sled at position 1. He leaves the ground at position 2 and lands in deep snow at a distance of b = 25 ft. How fast was he going at position 1?

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To find out how fast the boy was going at position 1, we need to use the principles of conservation of energy. The sled loses gravitational potential energy as the boy jumps, which is converted into kinetic energy. Therefore, we can equate the initial gravitational potential energy to the final kinetic energy.

First, we need to determine the gravitational potential energy at position 1. Gravitational potential energy is given by the equation:

PE = mgh,

where m is the mass (given as m), g is the acceleration due to gravity, and h is the height from the ground to position 1.

Since we are given that the mass of the boy is m and he is on the sled, we can assume the sled's mass is negligible. Let's assume that position 2 is the same height as the ground, so h is the height from the ground to position 1. We also know that h equals the total distance the boy travels, b (25 ft).

Therefore, the gravitational potential energy at position 1 is:

PE1 = mgh.

Next, we need to determine the kinetic energy when the boy lands in the deep snow at position 2. Kinetic energy is given by the equation:

KE = 0.5mv^2,

where m is the mass (given as m) and v is the velocity.

Since the sled lands in the deep snow, we can assume that there is no further height change, which means the final kinetic energy is equal to the potential energy lost. Therefore:

PE1 = KE2.

Substituting the equations for potential and kinetic energy, we get:

mgh = 0.5mv^2.

Simplifying the equation by canceling the mass and rearranging, we find:

v^2 = 2gh.

Finally, we can solve for v by taking the square root of both sides:

v = √(2gh).

Now we can plug in the known values. The acceleration due to gravity, g, is approximately 32 ft/s^2. The height from the ground to position 1, h, is equal to the distance traveled, b, which is given as 25 ft.

Plugging in these values, we get:

v = √(2 * 32 ft/s^2 * 25 ft).

Simplifying further, we find:

v ≈ √(1600 ft^2/s^2) ≈ 40 ft/s.

Therefore, the boy was going approximately 40 ft/s at position 1.