Rectilinear Motion (Object dropped 500m, will it burst?)

In summary: You can just use the quadratic formula to solve 500 = 4.9t^2 for t.You don't need to use the kinematic equations, but you could use them if you used s=500m, u=0, a=-9.8m/s^2, and solved for t.But you don't need to use any equations at all. Not really.If the object falls 500m, what is its final velocity?It starts from rest, so it is easy to figure out the final velocity. It has fallen 500m in t seconds. So you just need to find t so that the object falls
  • #1
01010011
48
0

Homework Statement



An object is dropped from a height 500 m above the ground. This object is designed to withstand an impact velocity of 100 m/s. Will it burst?

Homework Equations



gravitational force: 9.8 m/s

The Attempt at a Solution



s(t) = 500 m
a(t) = 9.8 m/s
v(t) = ?

a(t) = 9.8
Therefore, v(t) = 9.8t + C

What do I do from here? How do I find for C?
 
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  • #2
You have s=500 and a=-9.81, they tell you that it is dropped from that height (so what is the initial velocity u?)

What equation of motion relates, v with u,a and s ?
 
  • #3
I think you are on the right path.
v(t) = 9.8t + C
but you know something about the velocity at t = 0! This condition will give you C.

After you got v(t), you can get s(t) with the same method. Keep in mind the direction of v! Finally you can calculate the time for the object to hit the ground and its velocity.
 
  • #4
Thanks for your replies and guidance rock.freak667 and justsof.

Ok, here is another attempt based on your what you said:

Now the v = g*t

a(t) = -9.8 m/s
v(t) = -9.8t + C
v(0) = -9.8(0) + C = 0
C = 0

v(t) = -9.8t + 0

s(t) = (-9.8t^2)/2 + D

(-9.8t^2)/2 + D = 500

(-9.8(0)^2)/2 + D = 500

D = 500

(-9.8t^2)/2 + 500
I am really confused at this point
 
  • #5
Right so then you have s(t)=500-(9.8/2)t2

s(t) is measured from the ground (at 0 displacement) to the object. So you want to find the 't' at which the object hits the ground.

So you want s(t) to be equal to what value?
 
  • #6
rock.freak667 said:
Right so then you have s(t)=500-(9.8/2)t2

s(t) is measured from the ground (at 0 displacement) to the object. So you want to find the 't' at which the object hits the ground.

So you want s(t) to be equal to what value?

I want s(t) to be equal to 500 m? At this point, I am just guessing but...
 
  • #7
01010011 said:
I want s(t) to be equal to 500 m?

No s is measured from the ground to the object. So at the ground, shouldn't s=0?
 
  • #8
rock.freak667 said:
No s is measured from the ground to the object. So at the ground, shouldn't s=0?

Ok, yes you are right, the ground is 0, that makes sense,

But my question is, what rules should I follow when working out problems like this? My problem is, the only rules I know are: a(t) = v'(t) = s"(t), but I don't know when to apply them.

Ok, since the object has hit the ground now, we have the current location, so do we have to differentiate to get the velocity like this:

s(0) = 500 - (9.8(t)^2)/2
therefore v(t) = 19.6 This does not make any sense, I'll try again


or do we just plug in the 0 into the formula like this: s(0) = 500 - (9.8(0)^2)/2 = 500 This also does not make any sense to me

s(0) = 500 m
 
  • #9
0 = 500 - (9.8(t)^2)/2, what is the value of t?

then just use v=u+at.


But there was a much simpler way to do this question, where you did not have to go through finding the time.
 
  • #10
rock.freak667 said:
0 = 500 - (9.8(t)^2)/2, what is the value of t?

then just use v=u+at.


But there was a much simpler way to do this question, where you did not have to go through finding the time.

Isn't t = 0?
 
  • #11
the 't' you will use there is time taken for the object to fall 500m
 
  • #12
rock.freak667 said:
0 = 500 - (9.8(t)^2)/2, what is the value of t?

then just use v=u+at.

But there was a much simpler way to do this question, where you did not have to go through finding the time.

Thanks for your patience rock.freak667.

I obviously need to practice problems like this. Can you recommend any basic tutorials on line similar to this?

This is pressure!

Ok, I am I on the right track now:

0 = 500 - (9.8t^2)/2

0 = 250 - 9.8(t)^2

-250/-9.8 = t^2

square root of 25.5 = t

5 = t

Object drops at 5m/s ??
 
  • #13
01010011 said:
Thanks for your patience rock.freak667.

I obviously need to practice problems like this. Can you recommend any basic tutorials on line similar to this?

This is pressure!

Ok, I am I on the right track now:

0 = 500 - (9.8t^2)/2

0 = 250 - 9.8(t)^2

-250/-9.8 = t^2

square root of 25.5 = t

5 = t

Object drops at 5m/s ??

you forgot to divide by the 2, 500-(9.8/2)t2=0, t is not 5.
 
  • #14
rock.freak667 said:
you forgot to divide by the 2, 500-(9.8/2)t2=0, t is not 5.

I'll try again:

0 = 500 - (9.8t^2) /2
0 = 500 - 4.9t^2
-500/4.9 = t^2
-102 = t^2
square root of -102 = t
10.09 = t

Now as you said, v=u+at
v = ?
U = 0 (initial velocity)
a = 9.8 m/s
t = 10.09 seconds

Therefore, v = 0 + 9.8(10.09) = 98.88 m/s
Is this correct?
 
Last edited:
  • #15
and thus the velocity at t=10.09 is ?
 
  • #16
rock.freak667 said:
and thus the velocity at t=10.09 is ?

Jezzzz, did I still get it wrong!

The question asks whether the can, which can withstand impact velocity of 100 m/s, will burst.

So isn't the answer no it won't burst because its velocity is 98.88 m/s (see working below)

0 = 500 - (9.8t^2) /2
0 = 500 - 4.9t^2
-500/4.9 = t^2
-102 = t^2
square root of -102 = t
10.09 = t

Now as you said, v=u+at (here is your formula)
v = ?
U = 0 (initial velocity)
a = 9.8 m/s
t = 10.09 seconds

Therefore, v = 0 + 9.8(10.09) = 98.88 m/s
Is this correct?
 
  • #17
Yes that is correct.

An alternative way was to just use a kinematic equation that involves v,u,g and s.

v2=u2+2g(s-s0)
 
  • #18
01010011 said:
Jezzzz, did I still get it wrong!

The question asks whether the can, which can withstand impact velocity of 100 m/s, will burst.

So isn't the answer no it won't burst because its velocity is 98.88 m/s (see working below)

0 = 500 - (9.8t^2) /2
0 = 500 - 4.9t^2
-500/4.9 = t^2
-102 = t^2
square root of -102 = t
10.09 = t

Now as you said, v=u+at (here is your formula)
v = ?
U = 0 (initial velocity)
a = 9.8 m/s
t = 10.09 seconds

Therefore, v = 0 + 9.8(10.09) = 98.88 m/s
Is this correct?

sqrt(-102) isn't 10.09. You moved stuff around wrong in doing the algebra. Just a sloppy mistake, right?
 
  • #19
I'm not no physics expert, but, can't you just solve for vf and if it is less then 100m/s then it will not burst...i solved for vf and got 98.99m/s...this shows that the object will not burst because the impact velocity is less then 100m/s
 

1. Will the object burst when dropped from a height of 500m?

It is highly likely that the object will burst when dropped from a height of 500m. This is due to the increasing air pressure and friction acting on the object as it falls through the air. The higher the height from which the object is dropped, the greater the impact force upon landing, which can cause the object to burst upon impact.

2. How does air resistance affect the object's motion?

Air resistance, also known as drag, is a force that opposes the motion of an object through the air. As the object falls, the force of air resistance increases, slowing down the object's acceleration. This means that the object will not continue to accelerate at a constant rate, but instead will eventually reach a terminal velocity where the force of air resistance is equal to the force of gravity.

3. What is the equation for calculating the object's velocity during rectilinear motion?

The equation for calculating velocity during rectilinear motion is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. This equation assumes that the object is experiencing constant acceleration, which may not be the case if air resistance is a significant factor.

4. How does the mass of the object affect its motion when dropped from a height of 500m?

The mass of the object will not directly affect its motion when dropped from a height of 500m. The force of gravity will act on the object equally, regardless of its mass, causing it to accelerate towards the ground at the same rate. However, a heavier object may experience a greater impact force upon landing, potentially causing it to burst upon impact.

5. Is there a way to decrease the likelihood of the object bursting when dropped from a height of 500m?

One way to decrease the likelihood of the object bursting when dropped from a height of 500m is to decrease the impact force upon landing. This can be achieved by either reducing the object's mass or increasing the amount of time it takes to land, such as by using a parachute or other form of air resistance. Additionally, the object's shape and material can also impact its likelihood of bursting upon impact.

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