Question

A body is released at certain hight. what will it distance cover in 4sec. (keep - g=9.8m/s²).​

Answer 1

Solution :

Given :-

A case of free falling motion of body is provided.

To Find :-

Distance covered by body in first 4s.

Concept :-

For a body falling freely under the action of gravity, g is taken positive.

For a body thrown vertically upward, g is taken negative.

When a body is just dropped, u = 0

For a freely falling body, Formula of distance covered by body is given by

$\underline{\boxed{\bf{\purple{\large{s=\dfrac{1}{2}gt^2}}}}}$

• s denotes distance
• g denotes acceleration due to gravity
• t denotes time

Calculation :

$\implies\bf\:s=\dfrac{1}{2}gt^2\\ \\ \implies\sf\:s=\dfrac{1}{2}\times 9.8\times (4)^2\\ \\ \implies\sf\:s=9.8\times 8\\ \\ \implies\underline{\boxed{\bf{\pink{s=78.4\:m}}}}$