A body is thrown vertically upward with initial speed v_0 =…

A body is thrown vertically upward with initial speed $v_0 = 10\ \mathrm{m/s}$. (a) How high does it rise? (b) After what time does it fall back to the ground? (c) With what velocity does it hit the ground?

Explanation of Method

Height, flight time and landing speed of an upward throw.

Step 1: (a) Maximum height

At the top the speed is zero. From $0 = v_0^2 - 2gh$: $h_{\max} = \frac{v_0^2}{2g} = \frac{100}{19.62}$.

\[h_{\max} \approx 5.1\ \mathrm{m}\]

Step 2: (b) Time of flight

The rise takes $t_1 = v_0/g \approx 1.02\ \mathrm{s}$; the fall from the same height takes just as long: $T = \frac{2v_0}{g} = \frac{20}{9.81}$.

\[T \approx 2.04\ \mathrm{s}\]

Step 3: (c) Velocity on landing

\[ v = v_0 - gT = 10 - 9.81 \cdot 2.04 = -10\ \mathrm{m/s}. \]

\[v = 10\ \mathrm{m/s} \ \text{downward}\]

$h\ (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $1.02$ $2.04$ $5.1$ $h_{\max} = \tfrac{v_0^2}{2g} \approx 5.1\ \mathrm{m},\ v = 0$ $v_0 = 10$ $10\ \mathrm{m/s}\ \downarrow$ $\text{up: } 1.02\ \mathrm{s}$ $\text{down: } 1.02\ \mathrm{s}$

Height against time: a symmetric parabola. The way up and the way down take the same time, and the body returns with the same speed it was thrown with.

Result

\[\boxed{\,h_{\max} \approx 5.1\ \mathrm{m}\,}\]

\[\boxed{\,T \approx 2.04\ \mathrm{s}\,}\]

\[\boxed{\,v = 10\ \mathrm{m/s} \ \text{downward}\,}\]

Remark

Note. Without air resistance the motion is symmetric: at every height the body has the same speed going up and coming down, and the upward and downward trips take the same time.