Answer to Question #240465 in Physics for munsyi

Question

A cannon shoots a bullet from the roof of a building from the position (in SI units):

𝑟̂0 = 9 𝑖̂+ 4 𝑗̂+ 15 𝑘̂

and with initial velocity (SI units):

𝑣̂o = 13 𝑖̂+ 22.5 𝑗̂+ 15 𝑘̂

where the z-axis points upward in the vertical direction and with origin at the ground

level. Assuming that the air resistance is negligible, find the maximum height reached

by the bullet and the position where it hits the ground

Answer 1

Find the magnitude of the initial velocity vector:

$$v_0=\sqrt{v_{0x}^2+v_{0y}^2+v_{oz}^2}=30\text{ m/s}.$$

Find the angle of launch above the horizontal xOy plane:

$$\theta=\arctan\frac{v_{0z}}{\sqrt{v_{0x^2}+v_{0y}^2}}=30°.$$

The maximum height reached:

$$h_\text{max}=\frac{v_0^2\sin^2\theta}{2g}=11.5\text{ m},$$

or 11.5+15=26.5 m above the point of launch.

The range, the position where it hit the ground:

$$R=\frac{v_0^2\sin(2\theta)}{g}=79.5\text{ m}.$$

The position:

$$\theta_x=\arctan\frac{v_{0y}}{v_{0x}}=60°.\\\space\\ R_x=R\cos\theta_x=39.8\text{ m},\\ R_y=R\sin\theta_x=68.8\text{ m}.\\\space\\ \vec r=(9+R_x)\hat i+(4+R_y)\hat j+15\hat k=\\=48.8\hat i+72.8\hat j+15\hat k.$$