Question

For a free falling body under earth's gravity, we write

Answer 1

For a free falling body under earth's gravity, we write

$v=u+g t$

$s=u t+\frac{1}{2} g t^2$

$v^2=u^2+2 g s .$

Here ' ${g}$ ' is acceleration due to gravity. Its value at sea-level and at $45^{\circ}$ latitude is approximately $9.8 {m} / {sec}^2$ or $980 {cm} / {sec}^2$ or $32 {ft} / {sec}^2$.