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Inga [223]
3 years ago
5

A stone is dropped from a cliff 64 feet above the ground. Answer the following, using -32 ft/sec2 as the acceleration due to gra

vity. Show all work and submit to D2L.
a. Find functions that represent the acceleration, velocity, and position of the stone above the ground at time t.
b. How long does it take the stone to reach the ground?
c. With what velocity does the stone hit the ground?
Physics
1 answer:
Scilla [17]3 years ago
4 0

Answer:

(a) v = 32 t

h = 16 t^2

g = 32 ft/s^2

(b) 64 ft/s

Explanation:

height, h = 64 feet

g = - 32 ft/s^2

(a) Let the time is t .

Let the velocity after time t is v.

Use first equation of motion

v = u + at

- v = 0 - 32 t

v = 32 t

Let the distance is h from the top.

Use second equation of motion

h = u t + 0.5 at^2 \\\\- h = 0 - 0.5\times 32 \times t^2\\\\h = 16 t^2

The acceleration is constant for entire motion.

(b) Let the velocity is v as it hits the ground. Use third equation of motion

v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 32\times 64\\\\v = 64 feet/s

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Air will break down (lose its insulating quality) and sparking will result if the field strength is increased to about 3 × 106 N
MrRissso [65]

Answer:

Acceleration,a=5.27\times 10^{17}\ m/s^2

Explanation:

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3 years ago
Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors a
mote1985 [20]

Answer:

E=0

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3 years ago
a rugby player passes the ball 5.34 m across the field, where it is caught at the same height as it left his hand. at what angle
MakcuM [25]

Answer:

31.035^{\circ}

Explanation:

x = Displacement in x direction = 5.34 m

t = Time taken to travel the displacement

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u = Initial velocity of ball = 7.7 m/s

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Displacement in x direction is given by

x=u\cos\theta t\\\Rightarrow t=\dfrac{5.34}{7.7 \cos\theta}

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y=u\sin\theta t-\dfrac{1}{2}gt^2\\\Rightarrow 0=7.7\sin\theta \dfrac{5.34}{7.7\cos\theta}-\dfrac{1}{2}\times 9.81 (\dfrac{5.34}{7.7\cos\theta})^2\\\Rightarrow 0=7.7\sin\theta-4.905\times \dfrac{5.34}{7.7\cos\theta}\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905\times 5.34\\\Rightarrow 0=7.7^2\dfrac{\sin2\theta}{2}-4.905\times 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905\times5.34\times 2\\\Rightarrow \sin2\theta=\dfrac{4.905\times 5.34\times 2}{7.7^2}\\\Rightarrow 2\theta=\sin^{-1}\dfrac{4.905\times 5.34\times 2}{7.7^2}

\Rightarrow \theta=\dfrac{62.07}{2}\\\Rightarrow \theta=31.035^{\circ}

The angle at which the ball was thrown is 31.035^{\circ}.

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3 years ago
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