PLEASE I NEED TO KNOW 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?

A) 5.5 m/s

B) 9.2 m/s

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

answer: he got 23

explanation: