A diver running 2.4 m/s dives out horizontally from the edge of a vertical cliff and 2.6 s later reaches the water below. How high was the cliff?
The horizontal distance does not change the time it takes to fall through the height of the cliff. It's there probably to throw you off, or there's a second part of the question.
What is important is the initial velocity is horizontal, so the vertical component, u, is zero.
vertical drop
S = ut+(1/2)gt²
= 0 + (1/2)(9.8)*2.6²
=?
To find the height of the cliff, we can use the equation of motion:
h = (1/2) * g * t^2
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.
In this case, the diver runs horizontally before diving, so the vertical distance traveled is the height of the cliff. Also, we are given the initial horizontal velocity (2.4 m/s) and time taken (2.6 s).
First, let's find the horizontal distance traveled by the diver using the equation:
d = v * t
where d is the horizontal distance and v is the initial horizontal velocity.
Substituting the given values, we get:
d = 2.4 m/s * 2.6 s
d = 6.24 m
Now, since the diver traveled horizontally before diving, the horizontal distance is equal to the distance traveled by the diver in the air, which is also the horizontal distance at the bottom of the cliff.
Using this horizontal distance, we can find the time at which the diver reaches the ground using the equation:
d = v * t
where d is the horizontal distance (6.24 m) and v is the horizontal velocity in the air (0 m/s because the diver is no longer moving horizontally).
Solving for t, we get:
6.24 m = 0 m/s * t
t = 0
Since there is no time when the diver reaches the ground, this means that the diver reached the water immediately after jumping from the cliff. Therefore, the vertical distance traveled (height of the cliff) is equal to the distance traveled in the air.
Using the equation of motion:
h = (1/2) * g * t^2
we substitute t = 2.6 s and g = 9.8 m/s^2:
h = (1/2) * 9.8 m/s^2 * (2.6 s)^2
h = (1/2) * 9.8 m/s^2 * 6.76 s^2
h = 33.16 m
Therefore, the height of the cliff is approximately 33.16 meters.
To determine the height of the cliff, we can use the kinematic equation that relates distance, velocity, time, and acceleration.
The equation we'll use is:
d = vt + (1/2)at^2
Where:
d = distance traveled
v = initial velocity
t = time
a = acceleration
In this case, the diver is moving horizontally, so the initial velocity in the y-direction (vertical) is 0 m/s. The only acceleration acting on the diver is due to gravity, which is approximately 9.8 m/s^2.
We need to find the vertical distance (height) traveled by the diver, so we substitute the values we know into the equation:
d = (0) * (2.6) + (1/2) * 9.8 * (2.6)^2
Simplifying this equation, we have:
d = (1/2) * 9.8 * (2.6)^2
d = 33.176 meters
Therefore, the height of the cliff is approximately 33.176 meters.