We are solving problems from a topic called one dimensional motion. The body has motion either along X axis or Y axis and we need not bother about the force acting on it in this chapter. Let us consider a problem to solve. Let it is thrown up with a certain velocity. It is given that during the upward journey, it has crossed two points of known separation with known velocities. We are interested in knowing the maximum height that the stone can reach from the point of projection, The problem is as shown in the diagram below.

**Solution**

We know that the second equation of motion gives relation between initial velocity,final velocity, displacement travelled by a body and acceleration. Using that data we can express acceleration due to gravity in terms of initial velocity and height between the two points. We can further write the equation for the maximum height and solve the problem as shown in the diagram below.

**Problem**

In the given problem one stone is dropped from a certain height. Simultaneously another stone is thrown up with a certain velocity and they two meet some where in the journey. That location from where the first stone is thrown up is given to us in the problem. We need to measure after how much time, the two stones are going to meet. The problem is as shown in the diagram below.

**Solution**

We can again use the same third equation of motion. We have the maximum height that the first body can travel. Using the data, we can measure the initial velocity in terms of the height. Let us assume that the two stones are going to meet at a point X from the ground. We shall write once again equation for the displacement and substitute the value of the initial velocity as shown in the diagram below. So we can express the time in terms of height as shown below.

**Problem**

A projectile is a body having two dimensional motion. It is moving both along X and Y axes simultaneously. The maximum velocity is the velocity with which it is projected up with a certain angle to the horizontal. Its maximum velocity is given in the problem and we are here to measure the range and maximum height of the projectile. The problem is as shown in the diagram below.

**Solution**

The velocity of the projectile keep changing as its position is changing. At the maximum height, it will have only horizontal velocity and its vertical component of the velocity becomes zero. Thus at the maximum height, velocity of the projectile is minimum.Range is the maximum horizontal distance that the body can travel and maximum height is the maximum vertical distance it can reach. We have derived equations for them and we can solve the problem as shown in the diagram below.

**Problem**

It is given in the problem, a player kicks a ball with a initial velocity as well as an angle to the horizontal. It is given that the ball experience a maximum horizontal range. There is another player at a separation on the field at a certain distance on the other side. The second player wish to catch the ball before it strikes the ground. We need to measure the speed that the second player shall have to catch the ball before it strikes the ground. The problem is as shown in the diagram below.

**Solution**

It is given in the problem that the range of the ball is maximum. It is possible when only when the ball is projected from the ground with an angle 45 degree. Thus we can measure the the maximum range. We can identify that the second player is at a distance of 16 meter from the ball's maximum range. He need to travel that distance before it hit's the ground. The time he has is same to the time of flight. Thus he has to run on the ground for 16 meter in the specified time with uniform velocity. The solution can be solved as shown in the diagram below.

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