E-mail address:

SOLUTION OF THE PROBLEM

SOLUTION OF THE PROBLEM

SOLUTION OF THE PROBLEM

We are in the reference system moving with the speed and direction of the wind with respect to the ground. This is a view from above at the moment t=0 when the ball begins to fly and the wind begins to blow.

At this moment the speed of ball has the following parallel to the ground components: is the component, which is orthogonal to the wall and is caused by the initial kick. Another component, which is parallel to the wall and to the ground, equals to V1, and is directed opposite to the wind's direction. That is, if the wind blows from right to the left with the speed V1 with respect to the ground and parallel to the wall, in the chosen system of reference moving with the wind everything acquires a speed component equal to V1 in magnitude but directed from left to the right.

The movement of the ball is affected only by the force of gravity and the force of air resistance. Now, we start looking for the key to unlock the solution. It is important to realize that the force of the air resistance is ALWAYS directed exactly opposite to the vector of the ball's speed. What can we say about this force? First of all, its magnitude is a very complex function of the air and ball characteristics, such as the temperature of the air, its density, size of the ball, etc. A cunning student should have guessed by now that since we don't know all this, we should NOT attempt to solve this problem with the help of Newton's second law. In this case, what do we have left? We know that in the initial moment of time the ball's vector of speed is directed in such a way as shown on the picture, view from above. In fact, the picture shows the vector of ball's speed projected on the ground as Vground. This projection of the full speed of the ball on the ground will be a complex function of time because we do not know how to calculate the Fair, the force of air resistance. Air resistance clearly affects the full speed of ball and its projection on the ground. Note, however, that the projection of Fair on the ground, denoted as Fground on the picture above, will always be opposite to Vground. The force of gravity affects the movement of the ball in the direction normal to the ground. Therefore, the force of gravity does not affect the magnitude or direction of Vground. Only air resistance force can do that.

This shows a way to a simple and powerful solution. We see now, that since the projections of the ball's speed and the force of air resistance on the ground are directed along the same line, the force of air resistance can NOT change the direction of the ball's movement (AS SEEN FROM ABOVE). Let's state this strictly: in the reference system moving with the wind the projection of the ball's trajectory on the ground will be a straight line directed along the projection of the ball's speed on the ground at moment of time t=0.

What is left, is geometry. The point D' on the picture above is the point, where the ball will strike the wall at the end of its flight. In reality, in the chosen system of reference the point D itself moves at speed 10 m/s from left to right. D' shows the position of D at the end of balls flight. Let's denote the time in flight . We know now that the ball will move along the straight line BD', if seen from above. The tangency of (see picture) is You should check that On the other hand, the same tangency at t=0 is .

Comparing two expressions for tangency of , we find .This is correct answer, without approximations and difficult mathematics. No calculus, just two lines of formulae and a page of physical reasoning. Of course, on the Admission Exam at MIPT where this problem was offered to 17 year old, the required explanation could be shorter. We just wanted to make this problem as clear as possible. #

If you noticed a typo select it and press Ctrl+Enter.

MIPT in social networks

soc-fb soc-tw soc-li soc-li
Яндекс.Метрика