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A mountain climber encounters a gap in an ice field. The opposite side of the gap is 2.75m lower, and is separated horizontally by a distance of 4.1m. To cross the crevasse, the climber gets a running start and jumps in the horizontal direction. (A) What is the minimum speed needed by the climber to safely cross the gap? (B) What is the climber's velocity right before landing?​

User Marcumka
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2 Answers

5 votes

Final answer:

To safely cross the gap in the ice field, the mountain climber needs a minimum speed of 2.75 m/s. Their velocity right before landing will also be 2.75 m/s.

Step-by-step explanation:

To safely cross the gap, the mountain climber needs to have a minimum speed that allows them to clear the horizontal distance and also compensate for the difference in height between the two sides of the gap. We can calculate the minimum speed using the principles of projectile motion. Since the climber jumps horizontally, the vertical motion can be ignored. The time it takes to cross the gap can be found using the equation: t = d/v, where t is the time, d is the distance, and v is the speed. With the given values, the time it takes to cross the gap is 4.1m/2.75m/s = 1.49s.

The minimum speed can be found using the equation: v = d/t, where v is the speed, d is the distance, and t is the time. Substituting the values, we get v = 4.1m/1.49s = 2.75 m/s. Therefore, the minimum speed needed by the climber to safely cross the gap is 2.75 m/s.

Right before landing, the climber's velocity will be the same as the initial velocity. Therefore, the climber's velocity right before landing will be 2.75 m/s.

User Ertebolle
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3 votes

Answer:

A) 5.5 m/s

B) 9.2 m/s

Step-by-step explanation:

A) Given in the y direction:

Δy = 2.75 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

2.75 m = (0 m/s) t + ½ (9.8 m/s²) t²

t ≈ 0.75 s

Given in the x direction:

Δx = 4.1 m

a = 0 m/s²

t = 0.75 s

Find: v₀

Δx = v₀ t + ½ at²

4.1 m = v₀ (0.75 s) + ½ (0 m/s²) (.75 s)²

v₀ ≈ 5.5 m/s

B) Find vᵧ.

v² = v₀² + 2aΔy

vᵧ² = (0 m/s)² + 2 (9.8 m/s²) (2.75 m)

vᵧ ≈ 7.34 m/s

Find v.

v² = vₓ² + vᵧ²

v² = (5.5 m/s)² + (7.34 m/s)²

v ≈ 9.2 m/s

User Ajay Bidari
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6.9k points