Question

An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 47.5 s after firing. What are the x- and y-coordinates of the shell where it explodes, relative to its firing point?

x =	________ m
y =	________

Answer 1

ANALYSIS:

My first step it to determine the time for the shell to reach its maximum height. If this time (for the shell to reach its maximum height) is less that 47.5 seconds, then it explodes on the mountain side on its way up. Conversely, if this time is greater than 47.5 seconds, then it explodes on its way down.

SOLUTION:

T = time for shell to reach maximum height = V(sinA)/g

where

V = launch velocity = 300 m/sec
A = angle of launch = 52 deg
g = acceleration due to gravity = 9.8 m/sec^2(constant)

Substituting values,

T = 300(sin 52)/9.8

T = 24.12 sec

This means that the shell explodes (47.5 - 24.12 = ) 23.38 seconds after reaching its maximum height.

X = the x-coordinate of the shell at explosion = 300(cos 52)(47.5)

X = 8773.18 meters

Ymax = V^2(sin^2 A)/2g

Ymax = (300^2)(sin^2 52)/(2*9.8)

Ymax = 2851.35

On its way down,

H = (1/2)(9.8)( 23.38)^2 = 2678.45

Therefore, upon explosion on its way down, the shell is

2851.35 - 2678.45 = 172.9 meters from the ground

Coordinates of the shell upon explosion:

X = 8773.18 meters and Y = 172.9 meters