A constant vertical force F = 10.81 N directed upward acts…

A constant vertical force $F = 10.81\ \mathrm{N}$ directed upward acts on a body of mass $m = 1\ \mathrm{kg}$, initially at rest, for $t = 10\ \mathrm{s}$. To what height does the body rise?

Explanation of Method

Height reached under a constant upward force.

Step 1: While the force acts

\[ a = \frac{F - mg}{m} = 10.81 - 9.81 = 1\ \mathrm{m/s^2}, \]

\[ h_1 = \tfrac12 a t^2 = 50\ \mathrm{m}, \]

\[ v_1 = at = 10\ \mathrm{m/s}. \]

\[h_1 = 50\ \mathrm{m} \ \text{when the force stops}\]

Step 2: After the force stops

The body is still moving up at 10 m/s and rises further like a vertical throw: $h_2 = \frac{v_1^2}{2g} = \frac{100}{19.62} \approx 5.1\ \mathrm{m}$.

\[h_{\max} = h_1 + h_2 \approx 55.1\ \mathrm{m}\]

$h\ (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $5$ $10$ $50$ $55.1$ $F \text{ acts: } a = 1\ \mathrm{m/s^2}$ $\text{coasting up}$ $11$

Height against time. While the force acts the body climbs 50 m; when it stops at 10 s the body is still moving up at 10 m/s and coasts another 5.1 m.

Result

\[\boxed{\,h_1 = 50\ \mathrm{m} \ \text{when the force stops}\,}\]

\[\boxed{\,h_{\max} = h_1 + h_2 \approx 55.1\ \mathrm{m}\,}\]

Remark

About the original solution. Its 50 m is the height at the moment the force stops acting. The body does not stop there: it coasts up another 5.1 m (for about 1 s), so the greatest height it reaches is about 55.1 m.