Answer:
19.62 m/s
34.335 m/s
1.75 m
Explanation:
v = Final velocity
u = Initial velocity = 0
a = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
t = Time taken = 2 s
From the equations of linear motion we have
[tex]v=u+at\\\Rightarrow v=0+9.81\times 2\\\Rightarrow v=19.62\ \text{m/s}[/tex]
The speed of the object after 2 seconds is 19.62 m/s
t = 3.5 s
[tex]v=u+at\\\Rightarrow v=0+9.81\times 3.5\\\Rightarrow v=34.335\ \text{m/s}[/tex]
The velocity of the object when it strikes the ground is 34.335 m/s.
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{34.335-0}{2\times 9.81}\\\Rightarrow s=1.75\ \text{m}[/tex]
THe height from which the object was dropped is 1.75 m.
