A body is thrown vertically upward with speed v_0 = 10 m/s…

A body is thrown vertically upward with speed $v_0 = 10\ \mathrm{m/s}$ from the top of a tower of height $H = 25\ \mathrm{m}$. (a) How long does it take to hit the ground? (b) With what velocity does it hit the ground?

Explanation of Method

A body thrown up from the top of a tower.

Step 1: (a) Time until it hits the ground

Take upward as positive, with the ground at $y = 0$: \[y(t) = H + v_0 t - \tfrac12 g t^2 = 0 \;\Longrightarrow\; 4.905\,t^2 - 10t - 25 = 0.\]

\[ t = \frac{10 + \sqrt{100 + 4 \cdot 4.905 \cdot 25}}{2 \cdot 4.905} = \frac{10 + 24.30}{9.81}. \]

\[t \approx 3.50\ \mathrm{s}\]

Step 2: (b) Velocity on impact

\[ v^2 = v_0^2 + 2gH = 100 + 490.5 \;\Longrightarrow\; v = \sqrt{590.5}. \]

\[v \approx 24.3\ \mathrm{m/s} \ \text{downward}\]

Check the original way (up, then down)

Up: $ t_1 = \tfrac{10}{9.81} = 1.02\ \mathrm{s} $ to the top at $ 25 + 5.10 = 30.1\ \mathrm{m} $

Down: $ t_2 = \sqrt{2 \cdot 30.1 / 9.81} = 2.48\ \mathrm{s} $; total $ 3.50\ \mathrm{s}\ \checkmark $; also $ v = g\,t_2 = 24.3\ \mathrm{m/s}\ \checkmark $

$h\ (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $1.02$ $2$ $3.5$ $25$ $30.1$ $\text{top: } 30.1\ \mathrm{m}$ $3.5\ \mathrm{s},\ 24.3\ \mathrm{m/s}\ \downarrow$ $\text{tower}$

Height against time. The body first rises 5.1 m above the tower, then falls 30.1 m to the ground.

Result

\[\boxed{\,t \approx 3.50\ \mathrm{s}\,}\]

\[\boxed{\,v \approx 24.3\ \mathrm{m/s} \ \text{downward}\,}\]

Remark

Note. The impact speed does not depend on whether the body was thrown up or down with 10 m/s: in both cases $v^2 = v_0^2 + 2gH$. Only the flight time differs.