Question

An object is made to fall from the top of a tower of height 78.4m
1) Find the distance travelled by it in the first second.
2) The time after which it strikes the ground.
3) The velocity with which it strikes the ground.
(take g = 9.8 ms-1)

Answer 1

Answer:

1) 4.9m 2) 4s 3) 39.2m/s

Explanation:

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Answer 2

Given:-

→Height of the tower(h) = 78.4m

→Acceleration due to gravity (g) = 9.8m/s²

To find:-

→Distance travelled by it in the 1st second

→The time after which it strikes the

ground

→The velocity with which it strikes the

ground

Solution:-

In this case:-

•Initial velocity of the object will be zero as it was initially at rest.

•Acceleration due to gravity(g) will be +9.8m/s² as the body is freely falling.

By using the 2nd equation of motion, we get:-

=> s = ut +1/2gt²

=> s = 0(t) + 1/2×9.8×(1)²

=> s = 4.9×1

=> s = 4.9m

We shall again use the 2nd equation of motion, to find the time taken:-

=> h = ut +1/2gt²

=> 78.4 = 0(t) + 1/2×9.8×t²

=> 78.4 = 4.9t²

=> t² = 78.4/4.9

=> t² = 16

=> t = √16

=> t = 4s

Now, we shall use the 3rd equation of motion to find the final velocity of the object:-

=> v²-u² = 2gh

=> v²-0 = 2(9.8)78.4

=> v² = 1536.64

=> v = √1536.64

=> v = 39.2m/s

Thus:-

•Distance travelled by it in the 1st second

is 4.9m.

•The time after which it srikes the

ground is 4s.

•The velocity with which it strikes the

ground is 39.2m/s.