Question

If an object is dropped from a tall building and hits the ground 3.0 s. later, what is the magnitude of the object's displacement in 1.0 s.; in 2.0 s.?

1.0 s. =

m

2.0 s. =
m

Answer 1

Find the velocity of the object after one second.

v = vo + at

v = (0 m/s) + (9.8 m/s^2)(1 s)

v = 9.8 m/s

Now, using that, you can find the displacement in that one second between 1 and 2.

d = vot + (1/2)at^2

d = (9.8 m/s)(1 s) + (1/2)(9.8 m/s^2)(1 s)^2

d = 14.7 m

Answer 2

Answer: In 1.0 s , Displacement = 4.9 m

In 2.0 s, Displacement = 19.6 m

Explanation: Using formula for equation of motion

`h=u* t + (1)/(2)* g* t^{^(2)}`

Initially , u= 0 m/s , t= 1.0 s and g=9.8
`(m)/(s^(2))`

Therefore
`h_1= 0 + (1)/(2)* 9.8* 1^(2) m = 4.9 m`

In Case t=2.0 s

`h_2= 0 + (1)/(2)* 9.8* 2^(2) m = 19.6 m`