A shell is fired at v_0 = 200 m/s and α = 45^° to the…
A shell is fired at $v_0 = 200\ \mathrm{m/s}$ and $\alpha = 45^\circ$ to the horizontal. For what time must its fuse be set so that it explodes at a height of $h = 10\ \mathrm{m}$ on the way down, before hitting the ground?
Explanation of Method
Setting a fuse to explode 10 m above the ground.
Step 1: Vertical motion
\[ y(t) = v_0\sin\alpha\,t - \tfrac12 g t^2, \]
\[ v_0\sin\alpha = 141.42\ \mathrm{m/s}. \]
The whole flight lasts $T = 2v_0\sin\alpha/g \approx 28.83\ \mathrm{s}$.
Step 2: When is the shell at 10 m?
\[ 4.905\,t^2 - 141.42\,t + 10 = 0, \]
\[ t = \frac{141.42 \pm \sqrt{141.42^2 - 196.2}}{9.81} = \frac{141.42 \pm 140.73}{9.81}. \]
The small root, 0.07 s, is on the way up; the explosion must happen on the way down, so we take the large root: $t_{\text{fuse}} \approx 28.76\ \mathrm{s} \ \ (\approx 0.07\ \mathrm{s} \text{ before impact})$.
Check
Near the ground the shell falls at about 141 m/s, so 10 m takes $ 10/141 \approx 0.07\ \mathrm{s} $: $ 28.83 - 0.07 = 28.76\ \mathrm{s}\ \checkmark $
Result
\[\boxed{\,t_{\text{fuse}} \approx 28.76\ \mathrm{s} \ \ (\approx 0.07\ \mathrm{s} \text{ before impact})\,}\]
Remark
About the original solution. It set up the equation but did not solve it, and the stated result (0.29 s before impact) is wrong: the correct margin is only about 0.07 s, so the fuse time is about 28.76 s. By symmetry, the 0.07 s is exactly how long the shell needs to rise the first 10 m after firing.