1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kipiarov [429]
2 years ago
6

A ball is gently dropped from a height of 30m. If its velocity increase uniform at the rate of 10m/s with what velocity will it

strikes the ground ? After what time will it strike the ground?​
Physics
1 answer:
jekas [21]2 years ago
3 0

Explanation:

<h3>Need to FinD :</h3>

  • The velocity of the ball with strikes the ground.
  • The time of the ball after which it strikes the ground.

\red{\frak{Given}} \begin{cases} & \sf {A\ ball\ is\ gently\ dropped\ from\ a\ height\ of\ {\pmb{\sf{30\ m}}}.} \\ & \sf {Its\ velocity\ increase\ at\ the\ rate\ of\ {\pmb{\sf{10\ m/s^2}}}.} \end{cases}

We know that,

  • The ball is dropped from the height of 30 m as stated in the question. Therefore, it will have a initial velocity of 0 m/s. The distance travelled by the ball will be 30 metres as the distance travelled by the ball is equal to the height of the tower and that is 30 metres.

So, by using the third equation of motion, we will find out the final velocity of the ball.

\rule{200}{3}

Therefore,

\sf \dashrightarrow {2as\ =\ v^2\ -\ u^2} \\ \\ \\ \sf \dashrightarrow {2 \times 10\ m/s^2 \times 30\ m\ =\ v^2\ -\ (0\ m/s)^2} \\ \\ \\ \sf \dashrightarrow {v^2\ =\ 600\ m^2/s^2} \\ \\ \\ \sf \dashrightarrow {v\ =\ \sqrt{600\ m^2s^{-2}}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{Final\ velocity\ (v)\ =\ 24.49\ m/s \approx 24.5\ m/s}}}}_{\sf \blue{\tiny{Required\ velocity\ of\ ball}}}}

∴ Hence, the required final velocity of the ball with which it strikes the ground is 24.5 m/s. Since, we know that the initial velocity of the ball is 0 m/s and the acceleration of the ball is 10 m/s². So, by using the first equation of motion, we will find out the time of the ball after which it stikes the ground.

\rule{200}{3}

Therefore,

\sf \dashrightarrow {v\ =\ u\ +\ at} \\ \\ \\ \sf \dashrightarrow {t\ =\ \dfrac{v\ -\ u}{a}} \\ \\ \\ \sf \dashrightarrow {t\ =\ \dfrac{24.5\ m/s}{10\ m/s^2}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{Time\ (t)\ =\ 2.45\ s}}}}_{\sf \blue{\tiny{Required\ time}}}}

∴ Hence, the required time of the ball after which it strikes the ground is 2.45 seconds.

You might be interested in
A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

6 0
4 years ago
One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing
Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

m = m_{1}=m_{2}

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

v_{1}=u_{1}+u_{2}

Put the value into the formula

v_{1}=2.2i-0.80i

v_{1}=1.4i\ m/s

Along Y-axis

0=m_{1}v_{1}+m_{2}v_{2}

m_{1}v_{1}=-m_{2}v_{2}

v_{1}=-v_{2}

Put the value into the formula

v_{1}=-1.36j\ m/s

Then the final speed of the first ball

v_{1}=\sqrt{(1.4)^2+(1.36)^2}

v_{1}=1.95\ m/s

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

\tan\theta=\dfrac{v_{2}}{v_{1}}

\tan\theta=\dfrac{-1.36}{1.4}

\theta=\tan^{-1}\dfrac{-1.36}{1.4}

\theta=-44.16^{\circ}

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0
3 years ago
A train starts from rest and travels for 5.0 s with a uniform acceleration of 1.5 m/s2. What is the final velocity of the train?
alexandr1967 [171]

Answer:

Final speed of the train is 7.5 m/s

Explanation:

It is given that,

Uniform acceleration of the train is, a = 1.5 m/s²

It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

v=u+at

Here, train starts from rest so, u = 0

v=0+1.5\ m/s^2\times 5\ s  

v = 7.5 m/s

So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.

7 0
3 years ago
Read 2 more answers
After turning off the television, you approach it. As you get close, but not touching it, the hairs on your arm start to stand u
Amanda [17]

The answer is D) induction.

6 0
3 years ago
Read 2 more answers
what is the average gravitational force of attraction between the earth and the sun? the earth averages a distance of about 150
Tanzania [10]

Answer:

B

Explanation:

Hhhhh

5 0
3 years ago
Other questions:
  • "two planets have the same mass, but planet a has 3 times the radius of planet
    15·1 answer
  • This diagram represents a top-down view of an experiment on a table. The 250g and 100g masses are falling and are pulling the bl
    6·2 answers
  • What change would occur in the kinetic energy of vapor if a container of vapor is placed at 0 Kelvin?
    8·1 answer
  • The catapults on the uss george h.w. bush can launch aircraft from rest to a speed of 150 mph over a distance of 270 feet. find
    10·1 answer
  • Describe an object that has decreasing kinetic energy
    14·2 answers
  • In which item is energy stored in the form of gravitational potential energy?
    13·2 answers
  • Plz answer fast the question
    8·1 answer
  • How do you use vectors daily without realizing it was a vector?<br><br> I need help ASAP !!!
    6·1 answer
  • Answer pls!!
    12·1 answer
  • The graph in Figure 4.15 shows the position and time for two runners in a race. Who has the faster speed, Robin or Joel? Explain
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!