Question

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

Answer 1

The answer is A.

Step-by-step explanation:

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Answer 2

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Step-by-step explanation:

a) We have equation of motion

S = ut + 0.5at²

Here u = 0, and a = g

S = 0.5gt²

Distance traveled during the first second ( t =1 )

S = 0.5 x 9.81 x 1² = 4.905 m

Distance traveled during the first second = 4.905 m.

b) We have equation of motion

v² = u² + 2as

Here u = 0, s= 75 m and a = g

v² = 0² + 2 x g x 75 = 150 x 9.81

v = 38.36 m/s

Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

75 = 0.5 x 9.81 x t²

t = 3.91 s

We need to find distance traveled last second

That is

S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

Distance traveled during the last second of motion before hitting the ground = 33.45 m