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netineya [11]
3 years ago
7

Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the gr

ound. When the balls collide, they are moving n opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur? (Use any variable or symbol stated above as necessary.) Both balls experience constant acceleration once they are �n flight. Choose a coordinate system with the origin at the ground and the upward direction positive. When the balls collide they are at the same height above the ground.
Physics
1 answer:
kramer3 years ago
3 0

Answer:

2h / 3

Explanation:

Let after time t , collision occurs

distance covered by A

= 1/2gt²

Let u be the initial velocity with which B was thrown upwards

distance by B

= ut - 1/2 gt²

total distance by A and B

h = ut - 1/2 gt² +  1/2gt²

h = ut

t = h / u

velocity of A after t

= gt

= g x h/u

velocity of B after t

= u - gt

= u - g x h/u

given

g x h/u = 2 (u - g x h/u)

2u² - 2gh = gh

u = \sqrt{\frac{3gh}{2} }

t = h / u

= \sqrt{\frac{2h}{3g} }

Distance covered by A

= 1/2 gt²

= 1/2 x 2h/3

= h/3

Height at which collision occured

= h - h/3

= 2h /3

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A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm
ipn [44]

To solve this problem it is necessary to apply the trigonometric ratios of the given velocity components.

If we make a graph of the velocity vectors in their respective velocities according to the given description we will have something similar to the attached graph.

The angle could be obtained from the components of the opposite leg and the adjacent leg so that

tan\theta = \frac{x}{y}

\theta = tan^{-1}(\frac{x}{y})

The opposite leg value (y) is 40cm / s and the adjacent leg (x) is 30cm / s

\theta = tan^{-1}(\frac{30}{40})

\theta = 36.87\°

Therefore the final direction that does the first ball is 36.87°

6 0
3 years ago
if the vessel in the sample problem accelerates fir 1.00 min, what will its speed be after that minute ?
LUCKY_DIMON [66]

Answers:

a) 154.08 m/s=554.68 km/h

b) 108 m/s=388.8 km/h

Explanation:

<u>The complete question is written below: </u>

<u></u>

<em>In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of 1.80 m/s^{2}, it would go from rest to its top speed in 85.6 s.  </em>

<em>a) What was the speed of the vessel? </em>

<em> </em>

<em>b) If the vessel in the sample problem accelerates for 1.00 min, what will its speed be after that minute? </em>

<em></em>

<em>Calculate the answers in both meters per second and kilometers per hour</em>

<em></em>

a) The average acceleration a_{av} is expressed as:

a_{av}=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{\Delta t} (1)

Where:

a_{av}=1.80 m/s^{2}

\Delta V is the variation of velocity in a given time \Delta t, which is the difference between the final velocity V and the initial velocity V_{o}=0 (because it starts from rest).

\Delta t=85.6 s

Isolating V from (1):

V=a_{av}\Delta t + V_{o} (2)

V=(1.80 m/s^{2})(85.6 s) + 0 m/s (3)

V=154.08 \frac{m}{s} (4)

If 1 km=1000m and 1 h=3600 s then:

V=154.08 \frac{m}{s}=554.68 \frac{km}{h} (4)

b) Now we need to find the final velocity when \Delta t=1 min=60 s:

<em></em>

V=(1.80 m/s^{2})(60 s) + 0 m/s (5)

V=108 \frac{m}{s}=388.8 \frac{km}{h} (6)

5 0
3 years ago
A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular
bagirrra123 [75]

To solve this problem we will use the concepts related to energy conservation. Both potential energy, such as rotational and linear kinetic energy, must be conserved, and the gain in kinetic energy must be proportional to the loss in potential energy and vice versa. This is mathematically

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mgh = \frac{1}{2}mv^2 +\frac{1}{2} I\omega^2

Where,

m = mass

v = Tangential Velocity

\omega = Angular velocity

I = Moment of Inertia

g = Gravity

Replacing the value of Inertia in a Disk and rearranging to find h, we have

mgh = \frac{1}{2}mv^2 +\frac{1}{2} I\omega^2

mgh = \frac{1}{2}mr^2\omega^2 + \frac{1}{2}(\frac{1}{2}mr^2)\omega^2 )

h = \frac{3}{4} \frac{r^2\omega^2}{g}

Replacing,

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h = 5.607m

Therefore the height of the inclined plane is 5.6m

3 0
3 years ago
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