Question

A stone is dropped from the rest and Falls freely under gravity calculate the distance travel in the first 4 seconds neglect the resistance and considered G is equal to 9.8 MS square 2

Answer 1

A stone is dropped from the rest and falls freely under gravity. Also, acceleration due to gravity is 9.8 m/s².

We have to find the distance travel in the first 4 seconds.

First equation of motion:

v = u + at

Second equation of motion:

s = ut + 1/2 at²

Third equation of motion:

v² - u² = 2as

From above data, initial velocity (u) is 0m/s (as it is dropped from rest), time (t) is 4s and acceleration due to gravity (a = g) is 9.8m/s².

Using the Second Equation Of Motion i.e. s = ut + 1/2 at²

Substitute the known values in the above formula,

→ s = 0(2) + 1/2(9.8)(4)²

→ s = 0 + 4.9(16)

→ s = 0 + 78.4

→ s = 78.4

Therefore, the distance travelled in first 4 sec is 78.4 m.

Answer 2

Answer:

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