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
A mountain climber jumps a 2.4m -wide crevasse by leaping horizontally with a speed of 8.0...
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Doing test review. Checking to see if my answers correct.
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