A stone is dropped from the edge of a roof. (a) How long does it take to fall 4.9m? (b) How fast does it move at the end of that fall? (c) How fast does it move at the end of 7.9m? (d) What is its acceleration after 1s and after 2 s?

Explanation

(a) As the stone is dropped, its initial velocity,`u=0`,
` h=4.9m, a=g=9.8 m//s^(2)`, time `t=?`
From `h=ut+(1)/(2)g t^(2), 4.9=0+(1)/(2)xx9.8t^(2)=4.9t^(2) or t^(2)=4.9/4.9=1,t=sqrt1=1s`
(b) Finaly velocity,` v=?` at `t=1s`
From `v=u+g t, v=0+9.8xx1=9.8 m//s`
(c) Let v be Final velocity when`h=7.9m`.
From `v^(2) -u^(2) = 2 gh, v^(2) - 0 = 2 (9.8) 7.9`
or `v=sqrt(2xx9.8xx7.9)=sqrt(154.84)=12.4 m//s`
(d) The accceleration of a freely failling body remains the same at all times, i.e,.
`a=g=9.8 m//s^(2)`, after 1 s and after 2 s.