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

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  

In 1.0 s , Displacement = 4.9 m

               In 2.0 s, Displacement = 19.6 m

Explanation: Using formula for equation of motion

Initially , u= 0 m/s , t= 1.0 s and g=9.8

Therefore

In Case t=2.0 s

unbalanced force

explanation: