Question

A certain freely falling object, released from rest, requires 1.85 s to travel the last 27.0 m before it hits the ground.

(a) Find the velocity of the object when it is 27.0 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.)
m/s

(b) Find the total distance the object travels during the fall.
m

Answer 1

(a) We have to find v₁ (see the figure below):

The last 27 meters travelled before the fall will be noted with $\Delta y$, which has the formula:

$\Delta y=v_{1}\Delta t+\frac{g\left ( \Delta t \right )^{2}}{2}\: (1)$

(If I write the equation taking into account the sense of the axis as is illustrated in the figure, all the terms will be with minus. Thus, the equation is as was written above.)

$\Delta t=1.85\: s$$\Delta t=1.85\: s$

Extract v₁ from formula (1): $v_{1}=\frac{\Delta y}{\Delta t}-\frac{g\Delta t}{2}$

v₁~5.53 m/s

Taking into account the axis that was considered (see the figure), v₁ is negative.

(b) From the equation of velocity v₁ we extract the total time of movement (between point 0 and 2), T:

$v_{1}=v_{0}+g\left ( T-\Delta t \right )\Rightarrow T=\frac{v_{1}}{g}+\Delta t$

(v₀=0)

T~2.41 s

$H=v_{0}T+\frac{gT^{2}}{2}=\frac{gT^{2}}{2}\approx 28.54\: m$ (or 28.56m, depending on the precision when calculating v₁ and T)