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 49.5 s after firing. what are the x- and y-coordinates of the shell where it explodes, relative to its firing point?

The x- and y-coordinates are 9142.57 m and -304.425 m

Explanation:

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

For motion with constant acceleration, we know

Along the horizontal, x-axis, we might write this as

Measuring distances relative to the firing point means

we know that,

or,

By applying the values, we get,

The acceleration of gravity is vertically downward and is , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

we know, and , so,

                 y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

that the claims meet official rules.

don't misinterpret the claims