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

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

Initially , u= 0 m/s , t= 1.0 s and g=9.8 $\frac{m}{s^{2}}$

Therefore $h_1= 0 + \frac{1}{2}\times 9.8\times 1^{2} m = 4.9 m$

In Case t=2.0 s

$h_2= 0 + \frac{1}{2}\times 9.8\times 2^{2} m = 19.6 m$