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Bad White [126]

3 years ago

10

If a block of wood dropped from a tall building has attained a velocity of 78.4 m/s how long has it been falling

1 answer:

ryzh [129]3 years ago

5 0

Gravity adds 9.8 m/s to the speed of a falling object every second.

An object dropped from 'rest' (v = 0) reaches the speed of 78.4 m/s after falling for (78.4 / 9.8) = <em>8.0 seconds</em> .

<u>Note:</u>

In order to test this, you'd have to drop the object from a really high cell- tower, building, or helicopter. After falling for 8 seconds and reaching a speed of 78.4 m/s, it has fallen 313.6 meters (1,029 feet) straight down.

The flat roof of the Aon Center . . . the 3rd highest building in Chicago, where I used to work when it was the Amoco Corporation Building . . . is 1,076 feet above the street.

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A 5kg wheel rolls off a flat roof of a 50 m tall building at 12m/s.

Olegator [25]

Explanation:

a) Given in the y direction (taking down to be positive):

Δy = 50 m

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Δy = v₀ t + ½ at²

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t = 3.2 s

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v₀ = 12 m/s

a = 0 m/s²

t = 3.2 s

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2 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

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How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab

tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec time to reach Vy = 0

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H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

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6 months ago