Question

A body falls vertically from a height of 19.6 metres with an initial velocity equal to zero then what tome will it take to travel
A) the first metres
B)the last metres
Neglect air resistance everywhere

Answer 1

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 meter:

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 meter:

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 traveled)= 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 )1/2 which 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 initial velocity for the last one meter journey.

So, 1 = 19.1*t + 4.9t2

4.9t2 + 19.1t - 1 = 0  

Using the quadratic formula, 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 meter is much less than the time taken by the body to cover the first meter because the object already has a velocity when it starts covering the last meter.

Answer 2

Explanation:

Height =19.6 m

u =0

To travel first 1m

s=ut+1/2gt^2

1=5t^2(taking g=10m/s^2)

t=√1/5=0.45............1

To travel 19.6m  

s=ut+1/2gt^2

19.6=5t^2

t=√19.6/5=1.97.............2

time taken to 18.6 m

s=ut+1/2gt^2

18.6=5t^2

t=√18.6/5=1.92.............3

Now time taken to travel the last m (time taken to travel full height -(minus) time taken to travel 1m less than the full height)

1.97-1.92=0.05