Worked exercise — Fall with initial velocity

Problem

A stone is thrown upward from $h_0 = 10\;\text{m}$ with velocity $v_0 = 5\;\text{m/s}$. Calculate: (a) maximum height above the ground; (b) total flight time until it hits the ground; (c) impact velocity.

Solution

Choice of axes. We take the $y$ axis pointing upward with the origin at ground level, so the acceleration is $a = -g$ and the equation of motion is:

$y(t) = h_0 + v_0\,t - \frac{1}{2}\,g\,t^2$

(a) Maximum height. At the highest point the vertical velocity vanishes, $\dot y = v_0 - g\,t = 0$, so $t^* = v_0/g$. Substituting:

$h_\text{max} = h_0 + \frac{v_0^2}{2g} = 10 + \frac{25}{19{,}62} \approx 11{,}27\;\text{m}$

$\ev{h_\text{max} \approx 11{,}3\;\mathrm{m}}$

(b) Total flight time. We impose $y(t) = 0$, i.e. arrival at the ground:

$4{,}905\,t^2 - 5\,t - 10 = 0 \;\Rightarrow\; t = \frac{5 + \sqrt{25 + 196{,}2}}{9{,}81} = \frac{5 + 14{,}87}{9{,}81}$

$\ev{t \approx 2{,}03\;\mathrm{s}}$

(the negative root is discarded, as it has no physical meaning).

(c) Impact velocity. From the velocity equation $v = v_0 - g\,t$:

$v_\text{imp} = 5 - 9{,}81\cdot 2{,}03 \approx -14{,}9\;\text{m/s}$

$\ev{\abs{v_\text{imp}} \approx 14{,}9\;\mathrm{m/s}}$

The negative sign indicates that at the moment of impact the stone is moving downward; its magnitude is $14{,}9\;\text{m/s}$.

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Topics: Cinematica Concepts: Caduta libera Skills: Scelta degli assi