Question

A mountain climber jumps a 3.2m -wide crevasse by leaping horizontally with a speed of 7.1m/s . If the climber's direction of motion on landing is -45?, what is the height difference between the two sides of the crevasse?

Answer 1

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 <= v0*sqrt(2*h/g)

v0 >= 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.