Answer:
6.0 m below the top of the cliff
Explanation:
We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation
[tex]v^2-u^2 = 2gd[/tex]
where
u = 0 (it starts from rest)
g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)
h = 24 m is the distance covered
Solving for h,
[tex]v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s[/tex]
So the ball thrown upward is launched with this initial velocity:
u = 21.7 m/s
From now on, we take instead upward as positive direction.
The vertical position of the ball dropped from the cliff at time t is
[tex]y_1 = h - \frac{1}{2}gt^2[/tex]
While the vertical position of the ball thrown upward is
[tex]y_2 = ut - \frac{1}{2}gt^2[/tex]
The two balls meet when
[tex]y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s[/tex]
So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is
[tex]y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m[/tex]
So the distance below the top of the cliff is
[tex]d=24.0 - 18.0 = 6.0 m[/tex]
