A batholith is an intrusive igneous rock formation that has been uplifted and exposed to earth's surface

2.86 A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight

Rashid [163]

Answer:

Distance Below the top = 6,02 m

Explanation:

To get the Final velocity of the first ball (that will be the intial velocity of the second) you need to solve the kinematic equation for Velocity:

(1) $V_{1f} =V_{1O} - gt$

As the ball is dropped from rest $V_{1O} = 0$ , so:

(2) $V_{1f} = - gt$

Note that the velocity is going to be negative as the ball is going down. To get the time it would take the ball to reach de base you can use the kinematic equation for position:

(3) $h_{1} = h_{1O} + V_{1O} *t - \frac{1}{2} gt^{2}$

We need the answer when h=0, and from the initial conditions $V_{1O} = 0, h_{1O} = 24m$ , you get:

(4) $24m = \frac{1}{2} gt_{1f}^{2}$

Solving for t, with $9,8\frac{m}{s^{2}}$ and :

(5) $t_{1f} = \sqrt{\frac{24m*2}{g} } = 2,21 s$

Replacing this time in (2), the final velocity is:

(6) $V_{1f} = -21,66 \frac{m}{s}$

So the initial velocity of ball 2 is equial to this but oppossite in direction so: $V_{2O}= -V_{1f} = 21,66 \frac{m}{s}$

The general position equation for ball 2 is (considering $h_{2O} = 0$ :

(7) $h_{2} = V_{2O} *t - \frac{1}{2} gt^{2}$

They cross paths when $h_{1}=h_{2}$ so:

(8) $h_{1O} - \frac{1}{2} gt^{2} = V_{2O} *t - \frac{1}{2} gt^{2}$

Rearranging:

(9) $t_{cross} =\frac{h_{1O}}{V_{2O} }$

Replacing values:

$t_{cross} = 1,108 s$

To get the absolute position replace $t_{cross}$ on equation (3) or (7):

$h_{cross} = 17,98 m$

To get it below the top of te cliff:

$h_{cross, from above} = 24 m - 17,98 m$

$h_{cross, from above} = 6,02 m$

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