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KatRina [158]
3 years ago
12

Q2) The position of an artillery, with a speed of projectile 650, which can fire in any direction above the horizontal plane and

whose direction of fire can be adjusted necessarily and sufficiently fast, is (t) = 30 x 10m +20 x 10'my. Besides that, the position of a target is determined as Fr(t) =(-12 x 10 m + 100). Thus at which instant of time must the artillery be fired in order that the target is terminated as earlier as possible? (Ignore the friction, the air resistance and assume that there is no Coriolis force)

Physics
1 answer:
Wittaler [7]3 years ago
5 0

Answer:

The minimum time to reach the target is 2156s

Explanation:

Check attachment

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You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
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Answer:

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Explanation:

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On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
How do i solve this?
kenny6666 [7]

Answer:

hmmm i dont know....

Explanation:

i just wanted free point. TANKS YOU SIR!!

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