Stone thrown upwards

Problem

A stone is thrown vertically upwards with initial speed $v_0 = 25\;\mathrm{m/s}$. Neglecting air resistance and using $g = 9{,}8\;\mathrm{m/s}^2$, determine the maximum height reached and the time taken to reach it.

Solution

Setup. We choose the vertical axis pointing upwards, with the origin at the launch point. During the ascent the stone is subject only to the gravitational acceleration, directed downwards, so $a = -g = -9{,}8\;\mathrm{m/s}^2$. The equations of uniformly accelerated motion are:

$v(t) = v_0 - g\,t, \qquad s(t) = v_0\,t - \tfrac{1}{2}\,g\,t^2$

Time of ascent. At the highest point the instantaneous speed vanishes ($v = 0$). Imposing this condition:

$0 = v_0 - g\,t \quad \Rightarrow \quad t = \frac{v_0}{g} = \frac{25}{9{,}8} \approx 2{,}55\;\mathrm{s}$

Maximum height. It is convenient to use the time-independent relation $v^2 = v_0^2 - 2g\,h$ with $v = 0$:

$h = \frac{v_0^2}{2g} = \frac{25^2}{2 \cdot 9{,}8} = \frac{625}{19{,}6} \approx 31{,}9\;\mathrm{m}$

The stone therefore rises to about $32\;\mathrm{m}$, taking a little over two and a half seconds.

$\ev{h \approx 31{,}9\;\mathrm{m}, \quad t \approx 2{,}55\;\mathrm{s}}$

Links

Topics: Cinematica Concepts: Caduta libera