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stellarik [79]
3 years ago
10

A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f

ired at an upward angle of 34.5° to the horizontal. Determine: (a) the time the ball is in the air: (b) the impact velocity: (c) the horizontal distance out from the base of the cliff that the ball strikes the water.​
Physics
1 answer:
wel3 years ago
5 0

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m​

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ  and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m​

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This question is two problems in one ... at least.  Ordinarily, I wouldn't
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I need to point out that there is an inconsistency in the question, and it's
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When the block is in water, the force of gravity on it and its apparent
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on you when you're standing on the beach.  The force of gravity on the
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If the force of gravity on the block is one million tons, then it's a million
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isn't asking for the force of gravity on the block.  It's asking for what the
block SEEMS to weigh when it's in water.

Now that we've gotten that out of the way, here's how we need to
untangle the question in order to answer it:

-- The weight of the block in water is

       (its weight OUT of water) minus (the buoyant force on it IN the water).

We'll need to find both of those.

-- In order to find its weight out of water, we need to find its mass.

-- In order to find its mass, we need to massage the given volume and density.

Let's begin:

                                               Density = (mass) / (volume)

Multiply each side by (volume):  Mass = (density) x (volume) =

                                                       (7,840 kg/m³) x (0.08 m³)

                                                            Mass = 627.2 kg.

Weight (out of water) = (mass) x (acceleration of gravity)

Acceleration of gravity on Earth = 9.8 m/s² .

Weight of the steel block out of water = (627.2 kg) x (9.8 m/s²) 

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Weight of the block in water =  (6,146.6 - 784) = 5,362.6 newtons
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