A shell leaves a cannon at an angle $\alpha = 45°$ to the horizontal with speed $v_0 = 100$ m/s. Find the maximum height and the flight time of the shell. Neglect air resistance; $g = 10$ m/s².
The vertical component of the speed: $v_{0y} = v_0 \sin 45° = 100 \cdot 0.707 \approx 70.7$ m/s.
Maximum height:
$$h = \frac{v_0^2 \sin^2 \alpha}{2g} = \frac{100^2 \cdot 0.5}{2 \cdot 10} = \frac{5000}{20} = 250 \text{ m}$$
Flight time (rise plus fall):
$$t = \frac{2 v_0 \sin \alpha}{g} = \frac{2 \cdot 100 \cdot 0.707}{10} \approx 14.1 \text{ s}$$
$h = 250$ m, $t \approx 14.1$ s