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To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 340 m/s at 56.0° above the horizontal. It explodes on the mountainside 35.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?

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Answer:

Horizontal displacement (
x-coordinate): approximately
6.65 * 10^(3)\; {\rm m}.

Vertical displacement (
y-coordinate): approximately
3.86 * 10^(3)\; {\rm m}.

Assumption:
g = 9.81\; {\rm m\cdot s^(-2)} and air resistance on the projectile is negligible.

Step-by-step explanation:

Under the assumptions, the projectile would travel at a constant horizontal velocity, while accelerating downward with a constant vertical acceleration of
a_(y) = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}.

The position of the projectile at the given time
t = 35.0\; {\rm s} after launch can be found in the following steps:

  • Find initial horizontal and vertical components of velocity,
    u_(x) and
    u_(y).
  • Apply the SUVAT equation
    x_(y) = (1/2)\, a_(y)\, t^(2) + u_(y)\, t to find the vertical displacement
    x_(y) (
    y-coordinate) of the projectile at the given time.
  • Since horizontal velocity of the projectile is constant, find the horizontal position (
    x-coordinate) of the projectile at the given time using the equation
    s = u_(x)\, t.

Given that the projectile was launched at
u = 340\; {\rm m\cdot s^(-1)} at
\theta = 56.0^(\circ) above the horizontal, the horizontal and vertical components of initial velocity would be:

Initial horizontal velocity:


\begin{aligned} u_(x) &= u\, \cos(\theta) \\ &= (340.0\; {\rm m\cdot s^(-1)})\, \cos(56.0^(\circ)) \\ &\approx 190.126\; {\rm m\cdot s^(-1)}\end{aligned}.

Initial vertical velocity:


\begin{aligned} u_(y) &= u\, \sin(\theta) \\ &= (340.0\; {\rm m\cdot s^(-1)})\, \sin(56.0^(\circ)) \\ &\approx 281.873\; {\rm m\cdot s^(-1)}\end{aligned}.

To find the vertical displacement (
y-coordinate) of the projectile at the given time, substitute
a_(y) = (-g) = (-9.81)\; {\rm m\cdot s^(-2)},
t = 35.0\; {\rm s}, and
u_(y) \approx 281.873\; {\rm m\cdot s^(-1)} into the SUVAT equation
x_(y) = (1/2)\, a_(y)\, t^(2) + u_(y)\, t :


\begin{aligned}x_(y) &= (1)/(2)\, a_(y)\, t^(2) + u_(y)\, t \\ &\approx \left[(1)/(2)\, (-9.81)\, (35.0)^(2) + (281.873)\, (35.0)\right]\; {\rm m} \\ &\approx 3.86 * 10^(3)\; \; {\rm m}\end{aligned}.

(Rounded to three significant figures.)

Since the horizontal velocity of the projectile is constant, horizontal displacement (coordinate) of the projectile can be found using the equation
s = u_(x)\, t:


\begin{aligned}s &= u_(x)\, t \\ &\approx (190.126)\, (35.0)\; {\rm m} \\ &\approx 6.65 * 10^(3)\; {\rm m}\end{aligned}.

(Rounded to three significant figures.)

User Bernardo Marques
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