Question

A mountain climber jumps a 1.9-m-wide crevasse by leaping horizontally with a speed of 9.0 m/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

ok well the answer i've got doesn't use the angle provided but im pretty sure it should work all the same.
so the variables are:
Horizontal: u=9 Vertical: u=0
s=1.9 s=?
a=0 a=-9.8
t=? t= the same as the horizontal
therefore,
s=ut
1.9=(9).t
t=0.21111 seconds
this means that
s(vertical)= 1/2at^2
= (0.5).(-9.8).(o.2111)^2
=-0.4368m
=0.4368 meters

Answer 2

First, let's see what we know:
\[Vix=9.0 m/s\]\[Vi y=0 m/s\]\[Ax=0 m/s^2 \]\[Ay=-9.8 m/s^2 \]The length of the crevasse doesn't matter in this particular question. Since the jumper lands at a -45 degree angle the final Vy must have the same magnitude as the final Vx. Now you have enough information to solve the problem using :
\[Vf^2 = V0^2+2ad\]Plugging in values:
\[(-9.0m/s)^2 =0m/s + 2(-9.8 m/s^2)d\]Solving for d you get a distance (height difference) of -4.13m.