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Ganezh [65]
3 years ago
12

You throw a ball from the balcony onto the court in the basketball arena. You release the ball at a height of 7.00 m above the c

ourt, with an initial velocity equal to 9.00 m/s at 33.0° above the horizontal. A friend of yours, standing on the court 11.0 m from the point directly beneath you, waits for a period of time after you release the ball and then begins to move directly away from you at an acceleration of 1.80 m/s2. (She can only do this for a short period of time!) If you throw the ball in a line with her, how long after you release the ball should she wait to start running directly away from you so that she’ll catch the ball exactly 1.00 m above the floor of the court?

Physics
1 answer:
Mariana [72]3 years ago
6 0

Answer:

Your friend has to wait 0.26 s after you throw the ball to start running.

Explanation:

The equation that gives the position vector of the ball is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal positon

v0 = initial velocity

t = time

α = throwing angle

y0 = initial vertical position

g = acceleration due to gravity

The equation of displacement of your friend is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of your friend at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.

Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:

y = y0 + v0 · t ·sin α + 1/2 · g · t²

-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m

Solving the quadratic equation:

t = 1.71 s

Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):

x= x0 + v0 · t · cos α

x = 0m + 9.00 m/s · 1.71 s · cos 33°

x = 12.9 m

Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.

Let´s see, how much time it takes your friend to run that distance:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0, v0 = 0)

x = 1/2 · a · t²

1.9 m = 1/2 · 1.80 m/s² · t²

Solving for t

t = 1.45 s

Then, since the time of flight of the ball is 1.71 s, your friend has to wait

1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.

You might be interested in
Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec
sammy [17]

Answer:

1) The maximum jump height is reached at A. 0.337s

2) The maximum center of mass height off of the ground is B. 1.64m

3) The time of flight is C. 0.834s

4) The distance of jump is B. 7.49m

Explanation:

First of all we need to decompose velocity in its rectangular components, so

v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s

1) We use, v_{fy}=v_{iy}-gt, as we clear it for t and using the fact that v_{fy}=0 at max height, we obtain t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s

2) We can use the formula y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2} for t=0.337s, so

y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m

3) We can use the formula y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find total time of fligth, so 0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66, as it is a second-grade polynomial, we find that its positive root is t=0.834s

4) Finally, we use x=v_{x}t=8.05m/s(0.834s)=6.71m, as it has an additional displacement of 0.77m due the leg extension we obtain,

x=6.71m+0.77m=7.48m, aprox x=7.49m

5 0
4 years ago
Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin
Vedmedyk [2.9K]

Answer:

The correct answer is 2.8*10^{-5}ms^{-1}

Explanation:

The formula for the electron drift speed is given as follows,

u=I/nAq

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is 8.93/63.5 molcm^{-3}. Converting this number to m³ using very elementary unit conversion we get 140384molm^{-3}. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}

if we convert the area from mm³ to m³ we get A=80*10^{-6}m^{2}.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}

which is our final answer.

6 0
3 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be
Inessa05 [86]

Explanation:

The given data is as follows.

    Inner wheel Radius = 20 m,

   Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = \frac{V}{R}

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = \frac{u}{(R - d)}

                  u = V \times \frac{(R - d)}{R}

                    = V \times (1 - \frac{d}{R})

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = 2\pi rn = V \times (1 - \frac{d}{R})

Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

            = \frac{V}{2 \pi r \times (1 - \frac{0.75}{20})} (since 2d = 1.5m given, d = 0.75m),

             = \frac{V}{r} \times 0.1532

So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

                   = \frac{V}{r} \times 0.1651

5 0
3 years ago
Can a material have negative permittivity?
sleet_krkn [62]
No it can't it's material
3 0
3 years ago
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