A mountain climber jumps a 2.5 -wide crevasse by leaping horizontally with a spe

Question

A mountain climber jumps a 2.5 -wide crevasse by leaping horizontally with a speed of 7.4 . If the climber's direction of motion on landing is -45, what is the height difference between the two sides of the crevasse?

Explanation / Answer

Define coordinate system with the origin at the initial location of the jumper, the positive y axis pointing down, and the positive x axis in the direction of the jump. Neglecting air resistance, the jumper's horizontal position as a function of time is then given by: x(t) = v0*t The jumper's vertical position as a function of time is given by: y(t) = 0.5*g*t^2, where g is the acceleration due to gravity = 9.98 m/s^2 In order to clear the crevasse, the jumper must travel at least a distance W horizontally in the time it takes to fall a distance h vertically. The time it takes to fall a distance h can be found from setting y(t) = h and solving for t: h = 0.5*g*t^2 2*h/g = t^2 t = sqrt(2*h/g) Plug this into the equation for the horizontal distance and set x(t) >= w: w = sqrt(2*h/(g*w^2)) Until hitting the ground, the jumper's horizontal component of velocity remains equal to v0. The vertical component of the velocity as a function of time is given by: v_y(t) = g*t Assuming the jumper *just* clears the crevasse, at the moment of landing (t = sqrt(2*h/g)), we have that: v_y = g*sqrt(2*h/g) = sqrt(2*h*g) (We could also have gotten this result by considering conservation of energy -- the change in the jumper's gravitational potential energy must be equal to the change in kinetic energy) At the moment of impact, the jumper has a velocity of (v0, sqrt(2*h*g)). This is a velocity vector directed arctan((sqrt(2*h*g))/v0) = arctan(sqrt(2*h*g)/sqrt(2*h/(g*w^2)) = arctan(g*w) below the horizontal.

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