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 principles of projectile motion. Here's how you can approach this problem:
Step 1: Break down the initial velocity into its horizontal and vertical components.
The initial velocity of 10.5 m/s at an angle of 67.6° below the horizontal can be split into its vertical and horizontal components.
The vertical component (Vy) can be calculated as:
Vy = 10.5 * sin(67.6°)
The horizontal component (Vx) can be calculated as:
Vx = 10.5 * cos(67.6°)
Step 2: Determine the time it takes to travel across the crevasse.
Since there is no horizontal acceleration, the time it takes to travel across the crevasse is the same as the time it takes to land.
Use the vertical equation of motion to find the time (t):
Vertical displacement (Δy) = -4 m (since you land 4 meters lower)
Vy = g * t + v0y
Rearrange the equation and substitute the known values:
t = (Vy - v0y) / g
Step 3: Calculate the horizontal displacement (distance) traveled.
The horizontal displacement can be calculated using the formula:
Distance = Vx * t
Step 4: Determine the width of the crevasse.
The width of the crevasse is equal to the horizontal displacement.
So, now you can substitute the calculated values into the equation to find the width of the crevasse.
Does that help?