Daniel B. answered 09/27/21
A retired computer professional to teach math, physics
This problem requires a particular property of parabolas.
I do not know your math background, so let me just state it as a lemma with a proof
using calculus. If you did not have calculus, you will just need to believe me.
Lemma: A parabola given by y = kx² has slope s at the coordinates (s/2k, s²/4k).
Proof:
y' = 2kx
The constraint on the slope gives the equation:
2kx = s
Solving for x:
x = s/2k
Plugging the value of x into the equation of the parabola
y = k(s/2k)² = s²/4k
QED
Now for the physics.
Let
v = 7.6 m/s be the horizontal speed of the jump,
g = 9.81 m/s² be gravitational acceleration.
The trajectory of the climber will be a parabola resulting from a combination of two motions:
- A horizontal trajectory, which the climber would follow in the absence of gravity
- A downward fall, which the climber would suffer in the absence of any jump.
You can view the trajectory in an X-Y coordinate space, where
the X coordinate is horizontal, and
the Y coordinate is vertical, with origin being at the start point of the jump.
Then [x(t), y(t)] is the position of the climber after time t.
(We start measuring time t from the instant the climber leaves the surface.)
x(t) = vt
y(t) = -gt²/2,
The first equation gives
t = x(t)/v
Plugging into the second equation
y(t) = -gx²(t)/2v²
The resulting trajectory is a parabola expressed as a function of x:
y = -gx²/2v²
Now we can apply the lemma with
k = -g/2v²
s = -1 (which is the slope of -45°)
The lemma gives us landing coordinates
(v²/g, -v²/2g)
Plugging in actual values,
the climber lands at height
y = -7.6²/(2×9.81) = -2.9 m
and horizontal distance
x = 7.6²/9.81 = 5.8 m
This implies two things
- The climber made it to the other side and landed 3 m from the edge.
- The height difference is 2.9 m.