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enot [183]
3 years ago
9

A rock is thrown horizontally from the top of a radio tower lands 17.0 m from the base of the tower. if the speed at which the o

bject was projected was 9.50 m/s, how high is the tower?
Physics
1 answer:
VikaD [51]3 years ago
4 0
For this problem, we use the equations derived from the projectile motion. Particularly, we use the trajectory of the projectile equation written below:

y = xtanθ + gx/2v₀²cosθ
where
y is the vertical height
x is the horizontal distance
θ is the launch angle
g is 9.81 m/s²
v₀ is the initial velocity

Since it was thrown horizontally, θ = 0°. Substituting the values,

y = 17tan0° + (9.81)(17)/2(9.50)²cos0°
y = 0.924 m
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Answer:

C. less than 950 N.

Explanation:

Given that

Force in north direction F₁ = 500 N

Force in the northwest F₂ = 450 N

Lets take resultant force R

The angle between force = θ

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The resultant force R

R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}

R=\sqrt{500^2+450^2+2\times 450\times 500\times cos\theta}

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3 years ago
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Explanation:

Given:

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a = 9.8 m/s²

t = 4.7 s

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just olya [345]

Answer:

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Explanation:

given;

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find:

What is the range of the ball?

solution:

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Vo = 25 m/s

<u>consider x-motion using time of fight: x = Vox * t</u>

where x = R = range

t =<u> 2 Voy </u>

      g

R =<u> Vo² sin (2Ф)</u>

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plugin values into the formula:

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therefore, the range or the ball is 48.81 m

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Answer:

State A = piece of metal; State B = air

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