Question

A rock is dropped from a 110-m-high cliff. How long does it take to fall (a) the first 55.0 m and (b) the second 55.0 m?

Answer 1

Answer:

a) t = 3.35[s]; b) t = 1.386[s]

Explanation:

We can solve this problem by dividing it into two parts, for the first 55 [m] and then the second part with the remaining 55 [m].

We will take the initial velocity as zero, as the problem does not mention that the Rock was thrown at initial velocity.

And using kinematics equations:

$v_{f}^{2}= v_{o}^{2}+2*g*y\\where:\\v_{o}=0\\g=gravity = 9.81[m/s^2]\\y=55 [m]\\v_{f}^{2}=0+2*9.81*55\\v_{f}=\sqrt{2*9.81*55} \\v_{f}=32.85[m/s]$

Now we can calculate the time:

$v_{f}=v_{o}+g*t\\t=\frac{v_{f}-v_{o}}{g}\\ t=\frac{32.85-0}{9.81}\\ t=3.35[s]$

Now we can calculate the second time, but using as a initial velocity 32.85[m/s].

The final velocity will be:

$v_{f}^{2}= v_{o}^{2}+2*g*y\\v_{f}=\sqrt{v_{o}^{2}+2*g*y} \\v_{f}=\sqrt{32.85^{2}+2*9.81*55 } \\v_{f}=46.45[m/s]$

Now we can calculate the second time:

$t=\frac{46.45-32.85}{9.81} \\t= 1.386[s]$

Note: The reason the second time is shorter even though it is the same distance is that the acceleration of gravity increases the speed of the rock more and more as it falls.