Question

A gun fires a shell at an angle of elevation of 30° with a velocity of 2 x 10' ms! What are the horizontal and vertical components of the velocity? What is the range of the shell? How high will the hall rise?

Answer 1

$\begin{gathered} u=2\times10^3m/s^{} \\ at^{} \\ 30^{\circ} \end{gathered}$

The velocity of the horizontal component can be calculated below

$\begin{gathered} x=u\cos \text{ }\emptyset \\ V_x=2\times10^3\times\cos 30^{\circ} \\ V_x=2000\times\cos 30^{\circ} \\ V_x=1732.05080757\approx1732.1\text{ m/s} \end{gathered}$

The velocity of the vertical component can be calculated below

$\begin{gathered} V_y=u\sin \emptyset \\ V_y=2\times10^3\times\sin 30^{\circ} \\ V_y=1000\text{ m/s} \end{gathered}$

The range of shell(horizontal displacement) can be calculated below

$\begin{gathered} r=\frac{u^2\sin 2\emptyset}{g} \\ r=\frac{4000000\sin 2(30)}{9.8} \\ r=353479.756647\text{ } \\ r=353479.8\text{ m} \end{gathered}$

The height can be calculated below

$h_{peak}=\frac{1000^2}{2g}=\frac{1000000}{19.6}=51020.4081633=51020.41\text{ m}$