Ballistic curve

The soldier fired a ballistic projectile at a 45° angle. The first half of its path it ascended, and the second half it fell. How far and how high did it reach if its average speed was 1,200 km/h and 12 seconds elapsed from the shot to impact?

Final Answer:

x =  11.33 km
h =  176.58 m

Step-by-step explanation:

$$\begin{gathered} v=1200\text{km/h}\to \text{m/s}=1200:3.6\text{m/s}=333.33333\text{m/s} \\ \alpha =45{}^{\circ } \\ {t}_0=12\text{s} \\ g=9.81{\text{m/s}}^2 \\ {v}_x=v\cdot \cos \alpha =v\cdot \cos 45°=333.3333\cdot \cos 45°=333.3333\cdot 0.707107=235.70226\text{m/s} \\ {v}_y=v\cdot \sin \alpha =v\cdot \sin 45°=333.3333\cdot \sin 45°=333.3333\cdot 0.707107=235.70226\text{m/s} \\ x(t)=v_x\cdot t \\ y(t)=vt\sin \alpha -\frac{1}{2}g{t}^2 \\ y(t)=0 \\ vt\sin \alpha =\frac{1}{2}g{t}^2 \\ t=\frac{2\cdot v\cdot \sin \alpha }{g}=\frac{2\cdot v\cdot \sin 45°}{g}=\frac{2\cdot 333.3333\cdot \sin 45°}{9.81}=\frac{2\cdot 333.3333\cdot 0.707107}{9.81}=48.05347\text{s} \\ {x}_1=v_x\cdot t=235.7023\cdot 48.0535\doteq 11326.311\text{m} \\ x=x_1\to \text{km}=x_1:1000\text{km}=11326.311:1000\text{km}=11.33\text{km}=11.33\text{km} \end{gathered}$$

$$\begin{gathered} t_2=t_0/2=12/2=6\text{s} \\ h=\frac{1}{2}\cdot g\cdot t_2^2=\frac{1}{2}\cdot 9.81\cdot 6^2=176.58\text{m} \end{gathered}$$

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