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*10^3m/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*10^3*\cos 30^(\circ) \\ V_x=2000*\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*10^3*\sin 30^(\circ) \\ V_y=1000\text{ m/s} \end{gathered}`

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

`\begin{gathered} r=(u^2\sin 2\emptyset)/(g) \\ r=(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)=(1000^2)/(2g)=(1000000)/(19.6)=51020.4081633=51020.41\text{ m}`