A stone is dropped from the top of the tower of height h. After one second another stone is dropped from the balcony 20m below the top. both reach the ground simultaneously .So, what is the value of h?(take g=10m/s^{2})

Hi Kreeti

By the problem statement, you would have realized that since height h, g, and time are involved - therefore we might need to use the formula:

h = ut+(1/2)at^{2}

Take signs of all the values as positive downwards (do you understand what I mean by this important statement?).

Now since stones are dropped and not thrown, hence u will be = 0. We are now left with two unknowns - h and t. (Value of a comes from g).

Write the equation for the first stone which covers distance h in t seconds.

Write the equation for the second stone which covers distance (h - 20) in (t - 1) seconds.

Solve these two equations for the two unknowns and you will get your answer.

By the way, solving of these equations will involve some mathematical techniques (dealing with quadratic functions). Solve few problems like these to get accustomed.

The numericals on laws of motion are very simple. We have only 3 formulas. and total of 5 variables - u, v, a, t, s. If you notice, there are actually 2 formulas. The 3rd v^{2} = u^{2} + 2as can be derived by eliminating the variable t using first two formulas. (Yes, you can try it!)

Therefore to solve the problem:

1) Identify the knowns and unknowns

2) Write pertinent equations for each stage of motion or for each body under motion and try to relate them. e.g. For a car accelerating and then running with constant velocity - the final velocity of acceleration stage becomes initial velocity of constant velocity state. Similarly, in this numerical, the height covered and time taken for the second stone can be written in terms of height covered and time taken by the first stone.

3) While writing the equations, get the sign convention correct. viz. In which direction will you take all values as positive. (Note, time can never be negative)

4) Solve the equations for the required!

Wishes

Arpit

**The average speeds of a bicycle, an athlete and a car are 18km/h,7km/s and 2km/min respectively. Which of the three is the fastest and which is the slowest?**

hiii aadi!!!

My solution for your question is-

First of all we will convert the units . So,

Speed of the bicycle=18km/hr=5m/s

Speed of the athlete=7km/s=7000m/s (anyway not logically possible!)

Speed of the car=2km/min=100/3 m/s (or 33.33…m/s)

Now, its clear that the fastest is the athlete and the slowest is the bicycle.

Hope its correct.

Solution is correct, but I think he meant 7m/s for the athlete.

If indeed the athlete ran with a speed of 7 km/s, I think with good training he should train harder to achieve escape velocity of 11.2km/s, and be the next Superman!

Jokes apart, by the way, Escape velocity is the "scalar speed" on the surface of the earth required by a body to escape from the gravitational force of the earth. Calling it velocity is actually a misnomer, as fire a body in any direction (not towards the earth though!) with this speed and the body will escape the earth's gravitational force can go into limitless space.

Furthermore, talking of athletes, can anyone tell me about the world record of Men's 100m sprint? Who holds it, in which games, and what was the average speed of the athlete in m/s and km/hr?

Thank you sir for your advice .

According to your instructions , the solution that i worked out is

h=ut+1/2at^{2}…………………(1)

h-20=ut+1/2a(t-1)^{2}………(2)

As u=0,

h=1/2at^{2}

h-20=1/2a(t-1)^{2}

Substituting the value from the first equation

1/2at^{2}-20=1/2a(t-1)^{2}

1/2at^{2}-20=1/2a(t^{2}+1-2t)

1/2at^{2}-20=1/2at^{2}+1/2a-at

Eliminating1/2at^{2} from both sides.

-20=1/2a-at

Taking a=10

-20=a(1/2-t)

-20/10=-t+1/2

-2=-t+1/2

t=1/2+2

t=5/2sec

Now,

h=ut+1/2at^{2}

h=1/2*10*(5/2)^{2}

h=5*25/4

h=125/4

h=31.25m

Sir , please tell me whether my answer is right or wrong.

Correct Kreeti, well done!

I can see that you are very comfortable with numbers. Would like you to devise time saving tactics to reduce calculations as the next challenge for you.

Your solution is perfect, though you could have reduced the calculations… e.g.

After you wrote:

Substituting the value from the first equation

1/2at^{2}-20=1/2a(t-1)^{2}

Put value of a = 10 => 1/2a = 5

5t^{2} - 20 = 5(t-1)^{2}

Divide everything by 5 to make things easier:

t^{2} - 4 = (t-1)^{2}

t^{2} - 4 = t^{2} + 1 - 2t

-4 = 1 - 2t

2t = 5 => t = 2.5 s

You are going good, practice will make you better!

Arpit

**a ball is dropped from rest from a height of 12m. If the ball loses 25% of its kinetic energy on striking the ground, how high will it bounce?**

i will wait ..!

PANKAJ

IX-C

Hi Pankaj

In this problem, the ball loses its Kinetic Energy. Kinetic Energy is 1/2mv^{2}. Mass cannot be lost, therefore the ball loses its speed on hitting the ground. Putting new Kinetic Energy as 75 % (as 25% is lost) of Kinetic Energy before hitting the ground, calculate the new speed. We also know that after striking the ground the direction of motion will be upwards - you can calculate the height to which the ball will bounce.

For all the students on this forum:

Solving the problem won't take more than a minute, but I believe that the purpose of this forum is not only to solve problems but to instill clearer concepts and better problem solving skills in all of you.

Therefore, please do not wait for a solution and put down your understanding of the problem and your solution (incomplete or wrong - doesn't matter!).

Wishes

Arpit

P.S. Do tell me if you have studied Law of conservation of energy. If you have studied that (and understood!), the above problem should not take more than 5 seconds to solve!…

Let me not be too stringent!… It will take very little time… though not 5 seconds, which will require years of dealing with such equations!… That will require you all to concentrate on getting your concepts crystal clear and practicing problems.

Arpit

hello sir,

i have a numerical problem.

Q) the driver of a car travelling at 36 km/h .he applies the breakes to accelerate uniformly. the car stops in 10s. plot speed-time graph for this person. find the distance travelled by him.

Prashant singh

class- ix a