this is a physics problem dealing with motion in two dimensions?

Question

To start an avalanceh ona mountian slope, an artillery shell is fired with an initial velocity of 320 m/s at 47degrees above the horizontal. It explodes on the mountainside 36.0 seconds after firing. What are teh x adn y coordinates of the shell where it explode, relative to its firing point

Answer 1

Neglecting air resistance, the x-coordinate (i.e. horizontal position) of the shell as a function of time (t in seconds), relative to the firing point, is given by:

x(t) = 320 m/s * cos(47 deg) * t

(i.e., the x-component of the shell's velocity is simply 320*cos(47 deg) m/s, and that never changes because there are no external forces acting on the shell in the horizontal direction)

The y-coordinate of the shell as a function of time is given by:

y(t) = 320 m/s * sin(47 deg) * t + 0.5 * g * t^2

where g is the acceleration of gravity = -9.807 m/s^2 (note the negative sign).

Plug in t = 36 seconds and you'll find that:

x(36 sec) = 7856.621 meters
y(36 sec) = 2070.485 meters

Answer 2

A bullet can't travel for 36 seconds and start an avalanche -- let alone still be in the air -- under any normal conditions here on earth, at least not with its initial velocity and angle of firing. But if we were imagining that it would, I will need the atmosphere's psi at the firing point and the altitude (elevation) of the firing point.

Answer 3

in simple terms undergo in recommendations your kinematics equations, and undergo in recommendations that action interior the x course could be separated from the action interior the y course. (do not overlook to chop up your preliminary speed into x and y factors, additionally)