Time taken for body under free fall

In summary: Now try to translate that into words.In summary, a ball is dropped from the top of a cliff with a height of 90m. Using the equations T=√(2h/g), v=u-gt, and s=ut-gt2/2, we can determine that the time taken for the rest of the descent is 1.242 seconds. This is found by setting the final position to 0 and using the initial position of 45m and an initial velocity of 0m/s. This result shows that the total time for the ball's descent is 3 seconds, with half of the descent taking 1.242 seconds.
  • #1
Swetasuria
48
0

Homework Statement


A ball is dropped from the top of a cliff. If time taken for half of the descent is 3s, then what is the time taken for the rest of rest of the descent? Take g=10m/s2


Homework Equations


T=√(2h/g)
v=u-gt
s=ut-gt2/2

The Attempt at a Solution


T=√(2h/g)
h=T2g/2=45m

∴height of cliff=90m

v=u-gt=0-(10*3)=-30ms-1

s=ut-gt2/2
45=-30t-5t2
t=-3

:frown: Looks wrong to me. Time can't be negative, can it? Also, shouldn't the answer be less than 3?
 
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  • #2
Swetasuria said:
s=ut-gt2/2
45=-30t-5t2
t=-3
You're making a sign error. Use the sign convention consistently. If down is negative, what should be the sign of s?
 
  • #3
If the origin is on the ground, the distance (which is above the origin) will be positive, right?
 
  • #4
Swetasuria said:
If the origin is on the ground, the distance (which is above the origin) will be positive, right?
No. What you're calling 's' is really Δs, which is negative as the object is falling.

The full statement of position as a function of time would be (using y instead of s, for clarity):
y = y0 + ut - 1/2gt2

In your case, y = 0 (the final position, measured from the ground) and y0 = 45.

Let me know if that's clear.
 
  • #5
Okay. So y=yo+ut-gt2/2
0=45-30t-5t2
t=1.242s
 
  • #6
Swetasuria said:
Okay. So y=yo+ut-gt2/2
0=45-30t-5t2
t=1.242s
Looks good.
 

1. How does the time taken for a body under free fall change with height?

The time taken for a body under free fall does not change with height. This is because the time of free fall is determined by the acceleration due to gravity, which is constant regardless of the height at which the body is dropped.

2. Is the time taken for a body under free fall affected by air resistance?

Yes, the time taken for a body under free fall is affected by air resistance. As the body falls, it encounters air resistance, which slows down its acceleration. This results in a longer time of free fall compared to a body falling in a vacuum.

3. Does the mass of the body affect the time taken for free fall?

No, the mass of the body does not affect the time taken for free fall. This is because the acceleration due to gravity is independent of the mass of the body. Therefore, regardless of the mass, the time of free fall will be the same.

4. How does the time taken for free fall on the moon compare to that on Earth?

The time taken for free fall on the moon is longer than on Earth. This is because the moon has a weaker gravitational pull compared to Earth. Therefore, the acceleration due to gravity on the moon is lower, resulting in a longer time of free fall.

5. Can the time taken for free fall be calculated using a formula?

Yes, the time taken for free fall can be calculated using the formula t = √(2h/g), where t is the time, h is the height, and g is the acceleration due to gravity. This formula applies to objects falling in a vacuum, without air resistance.

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