Question

a stone is dropped from the height 4.9 m high roof find out the following a.time taken to fall 4.9m b.velocity just before it reaches the ground c. what is its acceleration after 1s and 2 s from start

Answer 1

(a)initial velocity(u)=0

g=9.8m/s^2

From Newtons Laws of Motion: h=ut-1/(2gt^2) {(-) sign indicates that the body is falling towards gravity}

now u=0

i.e; h=1/(2gt^2)

i.e;t^2=2h/g

i.e;t^2= (2*4.9)/9.8

i.e;t^2=1

so; t= 1 s
(b)

According to the equation of motion under gravity:
v² − u² = 2 gs
Where,
u = Initial velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 4.9 m
g = Acceleration due to gravity = 9.8 m s⁻²
∴ v² − 0² = 2 × 9.8 × 4.9
v² = 2 × 9.8 × 4.9 = (9.8)²
v = 9.8 m s^{− 1}
Hence, the velocity of the stone just before touching the ground C is

9.8 m s⁻¹_{.
(c)acceleration remain constant throught motion of stone
=acceleration due to gravity ,g=9.8m/s^2=acceleration after 1s nad 2s}