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2 years ago

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A drone is flying horizontally when it runs out of power and begins to free fall from 16 m. No drag. If it lands 40 m away (in t

he x-direction) from where it began to fall, what was its horizontal velocity while it was falling

1 answer:

marysya [2.9K]2 years ago

7 0

Answer:

the horizontal velocity while it was falling is 22.1 m/s.

Explanation:

Given;

height of fall, h = 16 m

horizontal distance, x = 40 m

The time to travel 16 m is calculated as;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 16}{9.8} } \\\\t = 1.81 \ s$

The horizontal velocity is calculated as;

$v_x = \frac{X}{t} \\\\v_x = \frac{40}{1.81} \\\\v_x = 22.1 \ m/s$

Therefore, the horizontal velocity while it was falling is 22.1 m/s.

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A basketball is shot from 2 meters up at an angle of 60° above the x axis at an initial velocity of 9 m/s. What is the maximum h

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Hope this helps a little

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initial velocity component up = 9 sin 60 = 7.79

v = 9 sin 60 - 9.8 t

when v = 0, we are there

9.8 t = 7.79

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Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

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Answer:

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I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

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so

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take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

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Answer:

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