An artillery shell is fired with an initial speed of 300m/s at 55 deegry about the horizontal, it explodes on a mountain side 42s after firing, what are the x and y coordinate of the shell where it explodes relative to its firing point?

x = 7226.94 m

y = 1677.4 m

Explanation:

This is a classic problem of a parabolic move. Let's analyze the given data.

The speed is 300 m/s but it's fired at 55°, so the x and y component of the speed would be:

Vx = 300 cos55 = 172.07 m/s

Vy = 300 sin55 = 245.75 m/s

Now, to get the x and y components of the shell, we need to apply the following expressions:

X = Xo + Vx*t    (1)

Y = Yo + Vy*t - 1/2 gt²   (2)

Xo and Yo is 0, and we already have the speed in x and y, so the components are:

X = 0 + (172.07)(42)

X = 7226.94 m

For the y component:

Y = 0 + (245.75)(42) - 1/2 (9.8)(42)²

Y = 10321.5 - 8643.6

Y = 1677.4 m

Hope this helps

Sorry can’t u just tryna get points

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.