Answer:
[tex]44.27m/s[/tex]
Explanation:
The kinematic equation
[tex]v_f^2=v_0^2+2ad[/tex]
gives the final velocity [tex]v_f[/tex] of the object given the initial velocity [tex]v_0[/tex], the acceleration [tex]a[/tex], and the distance traveled [tex]d[/tex].
For our case, the object is dropped; therefore,
[tex]v_0=0[/tex]
I.e. the initial velocity is zero. The acceleration due to gravity is
[tex]a=9.8m/s^2[/tex],
and the distance traveled is [tex]d=100m[/tex].
Putting the values into the equation we get:
[tex]v_f^2=0+2(9.8ms^{-2})(100m)\\\\v_f^2=1960\\\\v_f=\sqrt{1960} \\\\\boxed{v_f=44.27m/s}[/tex]
The final velocity of the object is 44.27 m/s.