Ballistic curve

The soldier fired the ballistic grenade at a 45° angle. The first half ascended, and the second fell. How far and height did it reach if his average speed was 1200 km/h, and what 12 seconds were taken from the shot to impact?

Correct answer:

x =   km
h =   m

Step-by-step explanation:

$$\begin{gathered} v=1200km/h=333.33333m/s \\ {v}_x=v\cos 45^{\circ }=235.7m/s \\ {v}_y=v\sin 45^{\circ }=235.7m/s \\ x(t)=v_{x}t \\ y(t)=vt\sin 45^{\circ }-\frac{1}{2}g{t}^2 \\ y(t)=0 \\ vt\sin 45^{\circ }=\frac{1}{2}g{t}^2 \\ t=\frac{2v\sin 45^{\circ }}{g}=48.0535s \\ x=v_x\cdot t=235.7\cdot 48.0535=\text{km} \end{gathered}$$

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