Calculating Cliff Height Using Acceleration Due to Gravity

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SUMMARY

The height of the cliff is calculated to be 54.5 meters using the kinematic equation s = ut + 1/2at², where the initial velocity (u) is 0 m/s, time (t) is 3.30 seconds, and acceleration (a) is 9.81 m/s², the acceleration due to gravity. The calculation involves substituting these values into the equation, resulting in s = 0 + 1/2 * 9.81 * (3.30)². This discussion emphasizes the importance of understanding the relationship between distance, velocity, time, and acceleration in solving basic kinematics problems.

PREREQUISITES
  • Understanding of basic kinematics principles
  • Familiarity with the kinematic equation s = ut + 1/2at²
  • Knowledge of gravitational acceleration (9.81 m/s²)
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study advanced kinematic equations and their applications
  • Learn about free fall and projectile motion in physics
  • Explore real-world applications of gravitational acceleration
  • Practice solving similar kinematics problems with varying parameters
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A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.30 s. How high is the cliff?
 
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Zz.
 
This is a basic kinematics problem so it works the same mathematically as say the "car problem".

In the car problem, you can figure out the horizontal distance something travels using time by the relationship:

X = Vi(t) + 1/2(A)t^2

There are 4 variables in this equation: distance, velocity, time, and acceleration. If you know three of them, you can solve for the missing one. So let's say the car was initially at rest. Your initial velocity would be Vi = 0. Let's say the car accelerates at a constant rate of 4 m/s. Then A=4 m/s. And let's say that you want to know how far the car traveled in say 20 seconds. Now you know T. Once you know three of the 4, you can plug these in and use Algebra to solve for the missing distance traveled, X.

Your problem is very similar to the car problem except instead of a car traveling horizontally you have a ball traveling in a straight line vertically. Without spoiling anything you are given the time, the initial velocity, and the acceleration in your problem as well. Think about it a bit and you should be able to figure out what these three values are. From there, you can use the equation and solve.
 
u = 0 -as it is dropped.
t = 3.3 secs.
height = s
a = acceleration due to gravity = 10 approx.
s = ut +1/2at^2 ( t^2 is t square i.e t to the power of 2)
s = 0 + 1/2 * 10 * 3.3*3.3
s = 5 * 10.9
s =54.5
Therefore the height of the cliff is 54.5 m.
 
I am still confused as to what exactly is the acceleration.I know the initial velocity is 0 and the time is 3.30 sec.But i don't know what the acceleration is!
 
Nevermind The acceleration is 9.80 thanks everyone for the help!
 
Well I hope you know what acceleration means. In this problem it equals to 9.81 m/s^2 (gravitation constant), and it is acceleration caused by gravity, because you are dropping the stone verticaly.
 

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