Question

guys pls solve today I have a very important exam ...

I will thank a lot of ur answers and mark u brainliest if I find helpful

pls... answer with steps I will also give free 50 points question later after the exam tomorrow​

Answer 1

Answer:

hope this will help you

Explanation:

Take the downward 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/s 2

(a) Time taken to travel the first metre:

In the case, s = 1m, u = 0m/s and g = 9.8 m/s 2

Use the formula s = ut + 1/2 at 2

Substituting the values, you get

1 = 0×1+(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 = 0.452s

(b) Time taken to travel the last metre:

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

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

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

g = 9.8bm/s 2

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 = 19.1 m/s

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

So, 1 = 19.1×t + 4.9t2

4.9t 2 + 19.1 t -1 = 0

Using the quadratic formula, t= 0.05, -3.9

Since the whole motion started at t = 0, t = ( - 4.9 ) is not valid.

Therefore, t = 0.05s.

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.

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