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2 years ago

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A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

for the cargo to hit the ground and the range it travels

1 answer:

OLEGan [10]2 years ago

7 0

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$

So the horizontal distance travelled is

$d=v_x t$

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

$d=(333.3)(22.6)=7533 m = 7.5 km$

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 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

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Answer:

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Explanation:

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a)

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$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

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which yields:

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$V_{x}=\frac{x}{t}$

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$x=V_{x}t$

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x=(1.5m/s)(0.495s)

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$a=\frac{v_{f}-v_{0}}{t}$

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$V_{yf}=(-9.8m/s^{2})(0.495s)$

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$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

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so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

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Answer:

Explained in explanation

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