You jump across and down a crevasse from rest. You land on the other side at a speed of 10.5 m/s directed 67.6° below the horizontal. The other side is 4 meters lower.
How wide was the crevasse?
No Idea if I am doing this right.
But so far what I done was
10.5Sin(67.6) = - gΔt
and got -9.71
To determine the width of the crevasse, you need to use the kinematic equations of motion.
First, let's express the given information in terms of known values:
- The initial speed (when you jump) is 0 m/s since you start from rest.
- The final speed (when you land) is 10.5 m/s.
- The angle below the horizontal is 67.6°.
- The vertical displacement is -4 meters (since it's 4 meters lower on the other side).
Now, let's use the following kinematic equations:
1. v_f = v_i + at (equation 1)
2. Δy = v_i*t + (1/2)at^2 (equation 2)
In equation 1, since the initial speed is zero, it simplifies to:
v_f = at (equation 1)
In equation 2, we can separate the vertical and horizontal components:
Δy (vertical displacement) = v_i(t) + (1/2)gt^2 (equation 2)
Now, let's solve for time (t) using equation 1:
10.5 m/s = a * t
Next, we can substitute this value of acceleration (a) into equation 2:
-4 m = 0*t + (1/2)*g*t^2
-4 m = (1/2)*g*t^2
To simplify, substitute the known value of acceleration due to gravity (g = 9.8 m/s^2) into the equation:
-4 m = (1/2)*9.8 m/s^2 * t^2
-4 m = 4.9 m/s^2 * t^2
Now, solve for time (t). Taking the square root of both sides,
t = sqrt((-4 m) / (4.9 m/s^2))
Calculating this gives you t ≈ 0.899 seconds.
Finally, substitute this value of time into equation 1 to find acceleration (a):
10.5 m/s = a * 0.899 s
Solving for acceleration (a),
a = (10.5 m/s) / (0.899 s)
Calculating this gives you a ≈ 11.68 m/s^2.
Now, let's find the horizontal displacement using the time (t) and acceleration (a):
Δx (horizontal displacement) = v_i * t + (1/2) * a * t^2
Δx = 0 m/s * 0.899 s + (1/2) * 11.68 m/s^2 * (0.899 s)^2
Calculating this gives you Δx ≈ 4.17 meters.
Therefore, the width of the crevasse is approximately 4.17 meters.