What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is L = 2.50{\rm m} wide and h = 9.50{\rm m} below the top of the cliff?
To determine the minimum speed at which she should leave the top of the cliff to miss the ledge at the bottom, we need to consider the horizontal and vertical distances involved.
Let's assume that when she leaves the top of the cliff, her velocity has a horizontal component (Vx) and a vertical component (Vy).
First, let's consider the horizontal distance. The horizontal distance she would travel during her fall is equal to the width of the ledge, which is 2.50 m.
To find the time it takes to travel this horizontal distance, we can use the formula:
time = distance / velocity
Since the vertical distance does not affect the horizontal motion, we can consider only the horizontal component of the initial velocity.
Next, let's consider the vertical distance. The vertical distance she would fall is equal to the height of the cliff minus the height of the ledge, which is 9.50 m.
To find the time it takes to fall this vertical distance, we can use the formula:
time = sqrt(2 * vertical distance / acceleration due to gravity)
The acceleration due to gravity, denoted as "g," is approximately 9.8 m/s^2.
Now, since the time it takes to fall vertically is the same as the time it takes to travel horizontally, we can equate the two time expressions:
distance / velocity = sqrt(2 * vertical distance / g)
Substituting the values we have:
2.50 / Vx = sqrt(2 * 9.50 / 9.8)
Now we can solve for Vx:
Vx = 2.50 / sqrt(2 * 9.50 / 9.8)
Vx ≈ 3.123 m/s
Therefore, her minimum speed (horizontal component) as she leaves the top of the cliff should be approximately 3.123 m/s in order to miss the ledge at the bottom.
how long to fall 9.5 m ?
h = 9.5
speed down at bottom, v = sqrt (2 g h)
= 13.65 m/s
so average speed down = 6.83 m/s
so time down = 9.5/6.83 = 1.39 second fall
u t = d
u ( 1.39) = 2.5
u = 1.8 m/s