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Natasha_Volkova [10]
3 years ago
12

A ball rolls off the end of a horizontal table that is 4 meters off the ground. It is measured that the ball lands 3 meters away

at the base. What speed did the ball roll off the end of the table? Sketch a picture to help. Show all work for full credit.

Physics
1 answer:
ElenaW [278]3 years ago
8 0

Answer:

The speed at which the ball rolled off the end of the table is 3.3 m/s

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.

The position vector of the ball can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity.

When the ball reaches the ground, its position will be:

r final = (3, -4)

Then:

3 = x0 + v0x · t

-4 = y0 + v0y · t + 1/2 · g · t²

Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:

3 m = v0x · t

-4 m = 1/2 · g · t²

We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:

-4 m = 1/2 · g · t²

-4 m = -1/2 · 9.8 m/s² · t²

8 m / 9.8 m/s² = t²

t = 0.9 s

Then:

3 m = v0x · 0.9s

3 m/ 0.9 s = v0x

v0x = 3.3 m/s

The speed at which the ball roll off the end of the table is 3.3 m/s

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Answer:

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3 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

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3 0
2 years ago
Joe and Max shake hands and say goodbye. Joe walks east 0.50 km to a coffee shop, and Max flags a cab and rides north 3.45 km to
timama [110]

Answer:

3.486 km

Explanation:

Suppose Joe and Max's directions are perfectly perpendicular (east vs north). We can calculate their distance at the destinations using Pythagorean theorem:

s = \sqrt{J^2 + M^2}

where J = 0.5 km and M= 3.45 km are the distances between Joe and Max to their original parting point, respectively. s is the distance between them.

s = \sqrt{0.5^2 + 3.45^2} = \sqrt{12.1525} = 3.486 km

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3 years ago
Jack and Jill are maneuvering a 3300 kg boat near a dock. Initially the boat's position is &lt; 2, 0, 3 &gt; m and its speed is
Paraphin [41]

Answer:

The workdone by Jack is  W_{jack} = -1050J

The workdone by Jill is  W_{Jill} = 0J

The final velocity is  v = 1.36 m/s

Explanation:

From the question we are given that

          The mass of the boat is m_b = 3300kg

          The initial position of the boat is   P_i  = (2 \r  i  + 0 \r j + 3\r k)m

           The Final position of the boat is  P_f = (4\r i + 0 \r j + 2\r k )\ m

           The Force exerted by Jack \r F = (-420\r i + 0 \r j + 210\r k) \ N

             The Force exerted by Jill  \r F_{Jill} =(180 \r i + 0\r j + 360\r k)

Now to obtain the displacement made we are to subtract the final position from the initial position

                                 \r P = P_f - P_i

                                    = (4\r i + 0\r j + 2 \r k) - (2\r i + 0\r j + 3\r k  )

                                     = (2\r i + 0\r j -\r k )m

Now that we have obtained the displacement we can obtain the Workdone

  which is mathematically represented as

                                                   W =\r  F * \r P

 The amount of workdone by jack would be

                                               W_{jack} =\r  F * \r P

                                                 = [(-420\r i +0\r j +210\r k)(2\r  i + 0\r j - \r k)]

                                                 = (-420) (2) + (210)(-1)

                                                = -840 - 210

                                               =-1050J

  The amount of workdone by Jill would be

                                                 W_{Jill} =\r  F * \r P

                                                        = [(180 \r i + 0\r j + 360\r k)(2\r i +0\r j -\r k)]

                                                       = (180 )(2) +(360)(-1)

                                                       =0J

According to work energy theorem the Workdone is equal to the kinetic energy of the boat

              W = K.E = \frac{1}{2} m *[v^2 - (1.1)^2]

             -1050  = 0.5*3300 [*v^2- (1.1)^2]

            -1050 = 1650 [v^2 -1.21]

               0.6363 = v^2 -1.21

                   v^2 = 0.6363+1.21

                    v^2 =1.846

                    v = 1.36\ m/s

                   

6 0
3 years ago
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