An artillery shell is fired with an initial velocity of 300m/s at 55.0 o above t

Question

An artillery shell is fired with an initial velocity of 300m/s at 55.0o above the horizontal. to clear an avalance,it explodes on a monuntainside 42.0 s after firing.
What are the x- andy-coordinates of the shell where it explodes, relative to istfiring point? An artillery shell is fired with an initial velocity of 300m/s at 55.0o above the horizontal. to clear an avalance,it explodes on a monuntainside 42.0 s after firing.
What are the x- andy-coordinates of the shell where it explodes, relative to istfiring point?

Explanation / Answer

vx = v cos = (300 m/s) cos 55

vx = (300 m/s) (0,574) = 172,1 m/s

vy = v sin = (300 m/s) sin 55

vy = (300 m/s) (0,819) = 245.7 m/s

For motion with constant acceleration, we know

s = so + vo t + (1/2) a t^2

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

x = xo + vxo t + (1/2) ax t^2

Measuring distances relative to the firing point means

xo = 0

We know that

ax = 0

or

vx = vxo = constant

vx = 172.1 m/s

x = (172.1 m/s) (42 s)

x = 72 282 m = 72.3 km

Along the vertical, y-axis, we might write this as

y = yo + vyo t + (1/2) ay t^2

Measuring distances relative to the firing point means

yo = 0

We know that

ay = - g = - 9.8 m/s^2

y = 0 + (245.7 m/s) t - (1/2) (9.8 m/s2) t^2

y = 0 + (245.7 m/s) (42 s) - (1/2) (9.8 m/s2) (42 s)^2

y = 1 675.8 m

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