You foolishly attempt to jump across a crevasse whose far side is W = 2.93 m away horizontally, and H = 1.35 m lower vertically. What minimum horizontal velocity must you have to make it to the other side? Assuming you have the minimum velocity you just calculated in the previous part of the question, what is your landing angle below the horizontal in degrees?

To calculate the minimum horizontal velocity required to reach the other side of the crevasse, we can use the principle of conservation of energy.

First, we need to calculate the potential energy difference between the initial and final points:

ΔPE = mgh

Where m is the mass of the object (assumed to be negligible for this calculation), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the vertical distance (H) the object needs to traverse.

ΔPE = mgh = (1.35 m) * (9.8 m/s²) = 13.23 J

Next, we can calculate the minimum kinetic energy required to travel the horizontal distance (W) using the formula:

KE = (1/2)mv²

Where v is the minimum horizontal velocity we are trying to find.

KE = (1/2)mv² = 13.23 J

Since mass (m) cancels out, we are left with:

(1/2)v² = 13.23 J

Simplifying further:

v² = 2 * 13.23 J = 26.46 J

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

v = sqrt(26.46 J) = 5.14 m/s

Therefore, the minimum horizontal velocity needed to make it to the other side of the crevasse is approximately 5.14 m/s.

To calculate the landing angle below the horizontal, we need to use trigonometry. We have the vertical distance (H) and the horizontal distance (W) traveled. The angle we want to calculate is θ.

tan(θ) = H / W

tan(θ) = 1.35 m / 2.93 m = 0.461

To find the angle θ, we can use the inverse tangent (arctan) function:

θ = arctan(0.461)

Using a calculator, we find that θ is approximately 24.93 degrees.

Therefore, the landing angle below the horizontal is approximately 24.93 degrees.