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3 years ago

2. A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal

1 answer:

8 0

Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels x after time t horizontally is

x = (7.2 m/s) t

The object's height y at time t is

y = 9.4 m - 1/2 gt²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is

v = -gt

(a) The object hits the ground when y = 0:

0 = 9.4 m - 1/2 gt²

t² = 2 * (9.4 m) / (9.80 m/s²)

t ≈ 1.92 s

at which time the object's vertical velocity is

v = -g (1.92 s) = -18.8 m/s ≈ -19 m/s

(b) See part (a); it takes the object about 1.9 s to reach the ground.

(c) The object travels a horizontal distance of

x = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m

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11. You want to calculate the displacement of an object thrown over a bridge. Using -10 m/s2 for acceleration due to gravity, wh

Natalka [10]

Displacement is d

Vf² = Vi² + 2 g d

(-20²) = (+10²) + 2 (-9.8) d

-19.6 d = 300

d = -15.3 m

negative means lower

time is t

d = Vi t + 1/2 g t²

-15.3 = 10 t + (-4.9) t²

4.9 t² - 10 t -15.3 = 0

t = 3.06 s

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3 years ago

jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

Greeley [361]

Answer:

C. 590 mph

$\vert v_{cj}\vert=589.49\ mph$

Explanation:

Given:

velocity of jet, $v_j=500\ mph$
direction of velocity of jet, east relative to the ground
velocity of Cessna, $v_c=150\ mph$
direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

$\vec v_j=500\ \hat i\ mph$

$\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)$

$\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph$

Now the vector of relative velocity of Cessna with respect to jet:

$\vec v_{cj}=\vec v_j-\vec v_c$

$\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )$

$\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph$

Now the magnitude of this velocity:

$\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}$

$\vert v_{cj}\vert=589.49\ mph$ is the relative velocity of Cessna with respect to the jet.

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