A body falls freely from rest. At point A its speed is v_A =…

A body falls freely from rest. At point A its speed is $v_A = 29.43\ \mathrm{m/s}$, and at a lower point B it is $v_B = 49.05\ \mathrm{m/s}$. Find the height difference between A and B and the time the body needs to get from A to B.

Explanation of Method

Height and time between two points of a free fall.

Step 1: Time from A to B

In free fall the speed grows by $g$ every second: \[t_{AB} = \frac{v_B - v_A}{g} = \frac{49.05 - 29.43}{9.81} = \frac{19.62}{9.81}.\]

\[t_{AB} = 2\ \mathrm{s}\]

Step 2: Height difference

\[ v_B^2 = v_A^2 + 2g\,h_{AB} \;\Longrightarrow\; h_{AB} = \frac{v_B^2 - v_A^2}{2g} = \frac{2405.9 - 866.1}{19.62}. \]

\[h_{AB} \approx 78.5\ \mathrm{m}\]

Check with the average speed

$ \tfrac{29.43 + 49.05}{2} \cdot 2\ \mathrm{s} = 39.24 \cdot 2 = 78.48\ \mathrm{m}\ \checkmark $

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $1$ $2$ $3$ $4$ $5$ $29.43$ $49.05$ $A$ $B$ $h_{AB} \approx 78.5\ \mathrm{m}$ $t_{AB} = 2\ \mathrm{s}$

In free fall the speed grows by 9.81 m/s every second, so $v_A = 3g$ and $v_B = 5g$: A is passed 3 s after release and B after 5 s. The shaded area is the distance between them.

Result

\[\boxed{\,t_{AB} = 2\ \mathrm{s}\,}\]

\[\boxed{\,h_{AB} \approx 78.5\ \mathrm{m}\,}\]

Remark

Note. The speeds are exactly $3g$ and $5g$: the body passes A 3 s after release and B 5 s after release, 44.1 m and 122.6 m below the start.