How Do You Calculate the Time and Speed of a Toy Car Falling Off a Table?

  • Thread starter Thread starter merceb49
  • Start date Start date
  • Tags Tags
    Falling Table
AI Thread Summary
To calculate the time it takes for a toy car to fall from a height of 1.325 m, the formula t = sqrt(2h/g) is used, where g is the acceleration due to gravity (approximately 9.8 m/s²). This results in a time of around 0.52 seconds for the fall. The horizontal distance traveled by the car is 0.425 m, leading to a horizontal speed of approximately 0.818 m/s. The discussion highlights confusion around the calculations and emphasizes the importance of using the correct height for vertical motion. Overall, the calculations confirm that the time and speed can be derived from the given parameters using appropriate physics formulas.
merceb49
Messages
17
Reaction score
0
A toy car runs off the edge of a table that is 1.325 m high. The car lands 0.425 m from the base of the table.

How long did it take to hit the ground?
How fast was the car going?

My tutor even had trouble with this, without a speed this is no solvable, at least to me. Suggestions on formulas? And you can't answer the second one without the first one, so any suggested formulas or words of wisdom?
 
Physics news on Phys.org
Calculate the time it takes the car to hit the ground first. Use the formula 0.5at^2 = distance. You know the distance the car fell is 1.325 m.

The speed of the car is simply (0.425m)/time.
 
So i do not have the time, or the acceleration. i can't calculate the acceleration because i don't have the mass of the car. I understand how to get the speed but I don't have the time so I don't know how to solve it. How can I find the acceleration?
 
I just dropped my pen, what was its acceleration while falling?
 
What's the acceleration due to gravity?
 
wow I'm stupid. but i tried this all
.5(9.8)t^2=1.325 and I got around 0.5. And this is still incorrect, should I not use the distance it fell horizontally or should I add both of the distances together?
 
merceb49 said:
wow I'm stupid. but i tried this all
.5(9.8)t^2=1.325 and I got around 0.5. And this is still incorrect, should I not use the distance it fell horizontally or should I add both of the distances together?

t=sqrt 2h/a
 
with that formula i got 0.3, assuming i did it correctly (t = sqrt of 2(.425)/9.8)
Is that not a ridiculous number?
 
merceb49 said:
with that formula i got 0.3, assuming i did it correctly (t = sqrt of 2(.425)/9.8)
Is that not a ridiculous number?

h=1.325, not .425
 
  • #10
Also, v_f^2 = v_i^2 +2ax
 
  • #11
sqrt of 2(1.325)/9.8=t=.5=fast?
 
  • #12
merceb49 said:
sqrt of 2(1.325)/9.8=t=.5=fast?

t = time, not velocity
 
  • #13
yeah I am trying to answer the first question, and i thank you for coping with my confusion... i understand the second question, i have to find how long it took to find the velocity but i still don't think .5 is the correct time
 
  • #14
a) v (vertical speed) = 0
vt+(1/2)*gt^2=s
t = sqrt (2s/g)

b)b) v = horizontal speed
vt+(1/2)at^2 = s
a = 0
v = s/t
 
  • #15
Does the answer key say otherwise? If not 0.520s is the correct time. Don't believe it, drop something from 1.3 m up and see how long it takes to fall to the ground. The initial velocity is only due to the distance the car moves in the horizontal position as there is no initial vertical velocity. The initial velocity, which is the horizontal velocity, is 0.818 m/s.

And get a better tutor.
 
  • #16
I don't have my calc. so i can't verify your answers, sorry.
 
  • #17
Bloodthunder said:
Does the answer key say otherwise? If not 0.520s is the correct time. Don't believe it, drop something from 1.3 m up and see how long it takes to fall to the ground. The initial velocity is only due to the distance the car moves in the horizontal position as there is no initial vertical velocity. The initial velocity, which is the horizontal velocity, is 0.818 m/s.

And get a better tutor.

Yes, you need a better tutor. I don't mind helping though.
 
  • #18
pr0blumz said:
I don't have my calc. so i can't verify your answers, sorry.

It should be right, I think, I did my calculations similar to what you posted earlier.
 
  • #19
i was keying it incorrectly, bloodthunder your correct. thank you for all the help on this problem
 
  • #20
merceb49 said:
i was keying it incorrectly, bloodthunder your correct. thank you for all the help on this problem

i guess your frustration got the best of ya.
 
Back
Top