A diver runs along a horizontal cliff top of a large lake. She makes contact with the water 1.40 s after leaving the cliff and her contact point is 3.2 m horizontally outward from the edge of the cliff.
a. What is her speed as she leaves the cliff?
b. How tall is the cliff?
c. What is her velocity as she makes contact with the water?
a. 3.2 m / 1.40 s
b. 1/2 * 9.8 m/s^2 * (1.40 s)^2
c. v^2 = (horizontal)^2 + (vertical)^2
... a. is horizontal
... vertical is ... 9.8 m/s^2 * 1.40 s
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To solve this problem, we can use the equations of motion and apply the principle of horizontal motion and vertical motion separately.
a. To find the diver's speed as she leaves the cliff, we need to determine the horizontal component of her velocity. Since there is no horizontal acceleration (assuming no air resistance), her horizontal velocity remains constant throughout the motion.
Given:
Time (t) = 1.40 s
Horizontal displacement (x) = 3.2 m
We can use the equation:
x = v_x * t
Solving for the horizontal velocity (v_x):
v_x = x / t
Substituting the given values:
v_x = 3.2 m / 1.40 s
v_x ≈ 2.29 m/s
Therefore, the diver's speed as she leaves the cliff is approximately 2.29 m/s.
b. To determine the height of the cliff, we can use the equation of vertical motion in freefall:
y = v_0y * t + (1/2) * g * t^2
Since the vertical displacement (y) is equal to the height of the cliff, and there is no initial vertical velocity (v_0y), we can rewrite the equation as:
h = (1/2) * g * t^2
where h is the height of the cliff and g is the acceleration due to gravity (-9.8 m/s^2).
Substituting the given value for time (t):
h = (1/2) * (-9.8 m/s^2) * (1.40 s)^2
h ≈ -9.85 m^2
Since the height cannot be negative, we take the absolute value of h:
h ≈ 9.85 m
Therefore, the cliff's height is approximately 9.85 meters.
c. To find the velocity of the diver when she makes contact with the water, we need to consider both the horizontal and vertical components of her velocity.
The horizontal component of her velocity remains constant at 2.29 m/s (as found in part a).
The vertical component of her velocity can be determined using the equation of motion:
v_y = v_0y + g * t
Since the initial vertical velocity (v_0y) is zero, the equation simplifies to:
v_y = g * t
Substituting the given value for time (t):
v_y = (-9.8 m/s^2) * (1.40 s)
v_y ≈ -13.72 m/s
The negative sign indicates that the velocity is directed downward. However, since we are interested in the magnitude of the velocity, we take the absolute value:
|v_y| ≈ 13.72 m/s
The velocity vector can be found by using the Pythagorean theorem:
v = √(v_x^2 + v_y^2)
Substituting the values for the horizontal and vertical components:
v = √(2.29^2 + 13.72^2)
v ≈ √(5.2441 + 188.7424)
v ≈ √193.9865
v ≈ 13.92 m/s
Therefore, the diver's velocity as she makes contact with the water is approximately 13.92 m/s.
To solve these questions, we can utilize the principles of kinematics, specifically the equations of motion. Let's break down each part of the question and find the answers step by step.
a. What is her speed as she leaves the cliff?
To find the speed, we can use the formula: speed = distance / time.
Given:
Time taken (t) = 1.40 s
Horizontal distance (s) = 3.2 m
The speed will be the horizontal component of the velocity, as there is no vertical information given.
So, speed = distance / time = 3.2 m / 1.40 s = 2.29 m/s.
Therefore, her speed as she leaves the cliff is 2.29 m/s.
b. How tall is the cliff?
To determine the height of the cliff, we need to calculate the vertical displacement of the diver. We can use the equation of motion for vertical displacement:
s = ut + (1/2)at^2
Where:
s = vertical displacement
u = initial vertical velocity (which is 0 as the diver starts from rest)
a = acceleration due to gravity (which is approximately 9.8 m/s^2)
t = time of flight (which is 1.40 seconds)
Since the diver starts from rest, the initial vertical velocity is 0. Therefore, we can simplify the equation to:
s = (1/2)at^2
Plugging in the values:
s = (1/2) * (9.8 m/s^2) * (1.40 s)^2
s = 9.8 * 0.98 m
s = 9.604 m
Therefore, the height of the cliff is approximately 9.604 m.
c. What is her velocity as she makes contact with the water?
To find the velocity, we can use the equation: velocity = distance / time.
Given:
Time taken (t) = 1.40 s
Horizontal distance (s) = 3.2 m
The velocity will have both horizontal and vertical components.
Horizontal velocity = distance / time = 3.2 m / 1.40 s = 2.29 m/s (as calculated in part a)
To find the vertical component of velocity, we can use the equation of motion:
v = u + at
Where:
v = final vertical velocity (which we need to find)
u = initial vertical velocity (which is 0 as the diver starts from rest)
a = acceleration due to gravity (which is approximately 9.8 m/s^2)
t = time of flight (which is 1.40 seconds)
Plugging in the values:
v = 0 + (9.8 m/s^2) * (1.40 s)
v = 13.72 m/s
Therefore, her velocity as she makes contact with the water is approximately 2.29 m/s horizontally and 13.72 m/s vertically.