To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 300mm/s?
To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 300meter per seconds at 55 degrees above the horizontal. It explodes on the mountain side 42 seconds after firing. What are the x and y coordinatesof the shell where it explodes, relative to its firing point?
I would like to figure out the time for the shell to reach its maximum height first.
Formula is
T = Vo(sin A)/g
where
T = time to reach maximum height
Vo = initial velocity = 300 m/sec.
A = angle of launch = 55 degrees
g = acceleration due to gravity = 9.8 m/sec^2
Substituting values,
T = 300 * sin 55/9.8
T = 25.08 sec.
This means that the shell exploded on its way down. It travelled another 16.92 sec (42 – 25.08) on the way down before exploding.
Maximum vertical height attained by shell = Vo^2(sin^2A)/2g
and substituting values,
H = (300^2)(sin^2 (55))/(2 * 9.8)
H = 3081.17 meters
So, coming down for 16.92 sec., the shell vertical distance from the ground is
S = 3081.17 – (1/2)(gT^2)
S = 3081.17 – (1/2)(9.8)(16.92^2)
S = 3081.17 – 1402.80
S = 1678.37 m
The horizontal displacement of the shell after 42 seconds = (300)*(cos 55)*(42) = 7227.06
ANSWER: After 42 seconds from being launched, the x and y coordinates of the shell (where the origin is its firing point) are
7227.06 meters and 1678.37 meters, respectively.
To solve this problem, analyze the horizontal (x) and vertical (y) directions separately.
Horizontally, we assume that there is no drag due to air, and so the velocity of the shell is constant until it hits the mountain. This means you simply do a distance = velocity*time equation, so x = (42s)(300m/s)(cos55). Note that the cos55 factor is there because they give you the velocity vector, you need to multiply by cos55 for the horizontal component of the velocity and by sin55 for the vertical component of the velocity.
Vertically, you need to use d = (v_i)(t) + 1/2(a)(t^2). Its initial velocity v_i is 300m/s * sin(55), the acceleration is simply g, and t is again 42 seconds, so y = (300 m/s)(sin55)(42s) + (1/2)(-9.8m/s^2)(42s)^2. Note it’s important that g is negative here; if we are looking for a positive y-coordinate relative to the initial position, we need to note that the acceleration is in the negative direction.
One thing you should notice, if you’re confused about how I’m applying these formulas, is that you can always use the formula d = (v_i)(t) + (1/2)(a)(t^2) as a complete description of projectile/freefall motion if you ignore drag. In the horizontal component I was indeed using this formula; I simply dropped out the second term because a = 0 (velocity is constant).
References :
I would like to figure out the time for the shell to reach its maximum height first.
Formula is
T = Vo(sin A)/g
where
T = time to reach maximum height
Vo = initial velocity = 300 m/sec.
A = angle of launch = 55 degrees
g = acceleration due to gravity = 9.8 m/sec^2
Substituting values,
T = 300 * sin 55/9.8
T = 25.08 sec.
This means that the shell exploded on its way down. It travelled another 16.92 sec (42 – 25.08) on the way down before exploding.
Maximum vertical height attained by shell = Vo^2(sin^2A)/2g
and substituting values,
H = (300^2)(sin^2 (55))/(2 * 9.8)
H = 3081.17 meters
So, coming down for 16.92 sec., the shell vertical distance from the ground is
S = 3081.17 – (1/2)(gT^2)
S = 3081.17 – (1/2)(9.8)(16.92^2)
S = 3081.17 – 1402.80
S = 1678.37 m
The horizontal displacement of the shell after 42 seconds = (300)*(cos 55)*(42) = 7227.06
ANSWER: After 42 seconds from being launched, the x and y coordinates of the shell (where the origin is its firing point) are
7227.06 meters and 1678.37 meters, respectively.
References :
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