A body of mass m = 160 g is thrown vertically upward at v_0…

A body of mass $m = 160\ \mathrm{g}$ is thrown vertically upward at $v_0 = 100\ \mathrm{m/s}$ and reaches its highest point after $t = 80\ \mathrm{s}$. What is the average air resistance force?

Explanation of Method

Average air resistance on a body thrown up.

Step 1: The data are inconsistent

On the way up, both gravity and air resistance point down, so the deceleration is at least $g$, and the body must stop within

\[ t \le \frac{v_0}{g} = \frac{100}{9.81} \approx 10.2\ \mathrm{s}. \]

A climb of 80 s would need a deceleration of only 1.25 m/s², less than $g$. That would require a force pushing the body up, not air resistance.

Step 2: With $t = 8\ \mathrm{s}$ (most likely intended)

\[ a = \frac{v_0}{t} = \frac{100}{8} = 12.5\ \mathrm{m/s^2}, \]

\[ mg + F_{\text{air}} = ma. \]

\[ F_{\text{air}} = m\,(a - g) = 0.16 \cdot (12.5 - 9.81). \]

\[F_{\text{air}} \approx 0.43\ \mathrm{N} \ \ (\text{for } t = 8\ \mathrm{s})\]

$v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $0$ $2$ $4$ $6$ $8$ $50$ $100$ $10.2$ $\text{with air: } 12.5\ \mathrm{m/s^2}$ $\text{vacuum: } g$ $\text{80 s would need } a = 1.25 < g$

Speed against time on the way up. Even without air the body stops after 10.2 s; air resistance only makes the deceleration larger and the climb shorter. The line for 8 s is drawn in red.

Result

\[\boxed{\,F_{\text{air}} \approx 0.43\ \mathrm{N} \ \ (\text{for } t = 8\ \mathrm{s})\,}\]

Remark

About the original solution. With 80 s it obtained $F_{\text{air}} = -1.37\ \mathrm{N}$ and dropped the minus sign. The minus sign means the force points upward, opposite to what air resistance does on the way up, so the result shows that 80 s is impossible. The time is most likely 8 s, which gives 0.43 N, about a quarter of the body's weight (1.57 N).