A body falls vertically from the height h= 19.6m with it's initial velocity equal to 0. What time will it take to travel: A) the first metre B) the last metre Neglect air resistance

Asked by  | 20th Apr, 2013, 08:27: PM

Expert Answer:

Take the downwards direction as positive.

You know the following:
Total height "h" = 19.6 m
Initial velocity "u"= m/s (supposing that time is measured in seconds)
Acceleration due to gravity "g" = 9.8 m/s2

a. Time taken to travel the first metre:

In this case, s = 1m, u = 0m/s and g = 9.8 m/s2

Use the formula s = ut + 1/2at2

Substituting the values, you get

1 = 0*t + (1/2)(9.8)t2
Therefore, 1 = 4.9 * t2
or
t=  (1/4.9)1/2 which is approximately equal to 0.452.
t =0.452 sec

b. Time taken to travel the last metre:

First, you have to find the velocity of the body when it is 1 m above the ground.

For this, you can use the formula v2 - u2 = 2as

u= 0 m/s, s (distance travelled)= 19.6 - 1 = 18.6 m and g = 9.8 m/s2

Therefore, v2 = 2*9.8*18.6
&v = ( 2*9.8*18.6 )1/2which is approximately equal to 19.1 m/s.
v=19.1m/s

Now, substitute this as "u" in s = ut + 1/2at2 because it is the intial velocity for the last one metre journey.

So, 1 = 19.1*t + 4.9t2

4.9t2 + 19.1t - 1 = 0 

Using the quadratic formuala, t = 0.05, -3.9. Since the whole motion started at t = 0, t = -3.9 is not valid. 

Therefore, t = 0.05 s.

You can notice that the time taken to cover the last metre is much less than the time taken by the body to cover the first metre because the object already has a velocity when it starts covering the last metre.

Answered by  | 22nd Apr, 2013, 11:40: AM

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