A mountain climber jumps a 2.4m -wide crevasse by leaping horizontally with a sp

Question

A mountain climber jumps a 2.4m -wide crevasse by leaping horizontally with a speed of 8.0 m/s. If the climber's direction of motion on landing is -45 degrees, what is the height difference between the two sides of the crevasse? B) where does the climber land? A mountain climber jumps a 2.4m -wide crevasse by leaping horizontally with a speed of 8.0 m/s. If the climber's direction of motion on landing is -45 degrees, what is the height difference between the two sides of the crevasse? B) where does the climber land? B) where does the climber land?

Explanation / Answer

As given in the question,

u(x) = 8 m/s and d = 2.4 m

Since, there is no acceleration in x-direction; so v(x) = u(x) = 8 m/s

where, v = speed of the climber when he reaches the other side of the creavasse.

As given in the question, this velocity is at 45 degree angle, so v(y) = v(x) = 8 m/s

The time taken for this jump,

t = d / u(x)  = 2.4 / 8 = 0.3 s

Using following equation for the y-directional motion,

s = u*t + (1/2)*a*t^2

=> h = u(y)*t + (1/2)*g*t^2 , where u(y) = 0

=> Height Difference = h = 0 + (1/2)*9.8*0.3^2 = 0.441 m

(B) The climber will land at the other side of the crevasse, which is at a horizontal distance of d = 2.4 m and at a depth of h = 0.441 m

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