Free Falling Object's Initial Height

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SUMMARY

The discussion centers on calculating the initial height from which a freely falling object descends, given that it takes 1.50 seconds to travel the last 30.0 meters before impact. Participants confirm the use of kinematic equations, specifically the equation for distance under constant acceleration: Distance (final) = Distance (initial) + (Velocity(initial) * time) + 0.5 * (Acceleration * time^2). The known quantities include acceleration (9.8 m/s²), final distance (0.00 m), and intermediate distance (30.0 m). The conversation concludes with a strategy to solve for both the initial height and the time elapsed to reach 30 meters.

PREREQUISITES
  • Understanding of kinematic equations
  • Knowledge of constant acceleration principles
  • Familiarity with basic physics concepts such as velocity and distance
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the kinematic equations in detail, focusing on their applications in free fall scenarios
  • Learn how to derive initial velocity from final velocity and time
  • Explore the relationship between distance, velocity, and time in physics
  • Practice solving problems involving free fall and constant acceleration
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of motion under gravity.

darkmagicianoc
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The problem reads: A freeley falling object requires 1.50s to travel the last 30.0 m before it hits the ground. From what height above the ground did it fall?

If anyone can please help me with this problem, I would be very grateful. Thank you in advance for your time and effort helping me with this problem!
 
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What equations might come in handy here? :smile:
 


I am thinking that the Kinematic equations might be useful, but I'm not sure if they could be used, or which ones.
 
darkmagicianoc said:
I am thinking that the Kinematic equations might be useful, but I'm not sure if they could be used, or which ones.

Well, how is, in general, initial height, initial velocity, time, acceleration and present height related?

Under the assumption of CONSTANT acceleration, that is..
 
Maybe using:

Distance (final) = Distance (initial) + [Velocity(initial)] (time) + .5(Acceleration)(time)^2 to get original velocity, and then subtract (9.8 m/s2)(1s) till hitting zero, n adding that time to the 1.5 s, and then try solving for distance?
 
First of all:

Make a list of the KNOWN quantities in that equation, and what is UNKOWN.
 
a = 9.8m/s2
xm = 30.0 m
tm = 1.50s
xf = 0.00
xi = ?

where i = initial, f = final, and m = somewhere in between
 
So, consider now the interval from the intermediate values to the final values, regarding the intermediate values as "initial" for this particular time, then the only unkown would be the velocity the object had 1.5 seconds prior to hitting the ground.

Agreed?
 
Yes, that makes sense.

So now that I know how to get the velocity (intermediate), Do I use that as the new Final velocity, put zero as the initial velocity, solve for time and then the distance (using 30.0 as the final distance) and then add the two distances together?
 
  • #10
:-p I think I know what to do now! Thank you for helping me!
 
  • #11
darkmagicianoc said:
Yes, that makes sense.

So now that I know how to get the velocity (intermediate), Do I use that as the new Final velocity, put zero as the initial velocity, solve for time and then the distance (using 30.0 as the final distance) and then add the two distances together?

Just one issue:
Remember in your second part of the problem (after you've found the velocity had at 30m height above the ground), you have TWO unkowns:

The initial height from which the object fell, and the time that has elapsed until you've reached a height of 30meters (and having achieved the solved for velocity).

Final note:
Instead of adding one equation, you might solve the second part by using the relation that relates distance&velocities&time, ignoring to solve explicitly for time.

Therefore, you'll need another equation as well, it is simplest to use the kinematic relation relating velocity&acceleration&time.
 
  • #12
:smile: Thank you for your help! It worked!
 

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