I to find the distance of the freefalling objects when one hits the ground.

  • Thread starter Thread starter Purpplehaze
  • Start date Start date
  • Tags Tags
    Ground
AI Thread Summary
Two objects are dropped from a height of 60 meters, one with an initial velocity of 20 m/s and the other with 15 m/s. The discussion focuses on using the kinematic equation D = iV * T + a*t^2/2 to calculate the time it takes for the first object to hit the ground. After substituting the values into the equation, a quadratic equation is formed, leading to a solution for time (T) of 2 seconds for the first object. The second object is then analyzed using the same method, resulting in a distance of 10 meters when the first object reaches the ground. The calculations assume no air resistance and simplify gravity to 10 m/s^2 for ease.
Purpplehaze
Messages
7
Reaction score
0

Homework Statement


The two objects are dropped from 60 m height. One is dropped having an initial velocity of 20m/s and another with 15 m/s. What is the distance of the objects when the first one hits the ground if there is no air resistance.


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org


What have you done so far? What kinematic equations describe the position of an object as a function of time?
 


D= iV * T + a*t^2/2 I think that's the formula to be used.
 


Purpplehaze said:
D= iV * T + a*t^2/2 I think that's the formula to be used.
Good. So how can you apply this to solve for the time it takes for the 20 m/s object to reach the ground?
 


Purpplehaze said:
D= iV * T + a*t^2/2 I think that's the formula to be used.

Plugin the values for first object and calculate t.
 


Would this work? 2(D-iV)/a = t
 


Purpplehaze said:
Would this work? 2(D-iV)/a = t

No, how did you get this?
 


Purpplehaze said:
Would this work? 2(D-iV)/a = t
No.

Done right, you'll end up with a quadratic equation.
 


Then how can I isolate t?
 
  • #10
  • #11


60=20t + 4.9t^2
 
  • #12


Purpplehaze said:
60=20t + 4.9t^2

Are you allowed to use g=10 m/s^2? That would make calculations a lot easier.
 
  • #13


You have two different objects, with two different initial velocities, each involving t^2. Once you put the initial velocities in, you will have two different quadratic equations. What are those equations?
 
  • #14


Purpplehaze said:
60=20t + 4.9t^2
Good. Now you can put it into standard form and solve it using the quadratic formula (or using whatever method you like).
 
  • #15


I got 2 as T for the first one.
 
  • #16


Purpplehaze said:
I got 2 as T for the first one.
Good!

Now see if you can figure out where the other object (the one thrown with the slower speed) is at that time. Use the same formula, but with different data.
 
  • #17


Ok I got the answer, the distance is 10.
Thanks a lot.
 
Back
Top