### Motion in One Dimension Problems with Solutions Six

One dimensional motion is the study of the motion of a body only along one direction. In the given problem, a boy sees a ball going up and then down through the window of known height. The total time of visibility the window is given to us in the problem. We need to measure the height above the window. The problem is as shown in the diagram below.

Equations of motion help us in understanding the motion of the one dimensional body. If any three parameters of them are known, by using he appropriate relation, we can find the forth physical quantity.

Solution

As the total time of the journey is one second the upward or downward journey takes half of that time and that is half second. As it is all ready  crossing a certain height h before window, by the time it reaches the window, it will acquire a velocity and we can find that velocity using the third equation of motion.

By substituting that data in the second equation of the motion, we can measure the height above the window as shown in the diagram below.

Problem

In this problem, we are dealing with a body thrown vertically upward with a known velocity. While it is crossing the top of the tower, a man tries to catch it and failed to catch it and it has gone  over crossed above him. Any way in the return journey after three seconds, the ball reached the same person. The boy is able to catch it in the return journey and we are here to measure the height of the tower where that man is standing. The problem is as shown in the diagram below.

Solution

As the velocity of vertical projection is given to us, we can measure the time of ascent. The man is at the maximum height of the tower and he tries to catch it. It is obvious that the tower is at a height less than the maximum height that the stone can travel. He failed to catch it there and it has gone above him further. It is given that it returned to the same place after three seconds. It means it has travelled up further 1.5 second and started returning back. Thus the time for which it has crossed the tower is only one second. By substituting that data in the height of the tower formula, we can find the height of the tower as shown in the diagram below.

Problem

In the given  problem, two stones are thrown vertically upward with the same velocity. But the second stone is thrown after three seconds of the first stone. We need to measure where do these two stones are going to meet ?

Solution

As the bodies are moving against the gravity, we need to treat the acceleration due to gravity as negative. As the times are different by three seconds, we need to take the second stone time as first stone time and less than three seconds as shown in the diagram below. But the height travelled by them has to be same so that they can meet at a certain height. By equating these equations, we can  get the time after which the two stones are going to meet and by further substituting that time data in any one of the height equation, we can measure the height of the meeting point as shown in the diagram below.

Problem

In the given problem, a velocity and time graph is given. We need to measure the ratio of distance travelled by the body in the first two seconds when compared with the total distance travelled by the body in the total time of the journey.

Solution

In the velocity and time graph, we can find the distance in the two seconds can be found as the area between the two seconds. Similarly we can also  find the the total distance as the total area of the graph in the total time as shown in the diagram below.

Problem

In this problem, a ball is thrown vertically up from the ground with a known velocity. At the same time ,another body is thrown down from a known height.It is given in the problem that after some time the two bodies got the collision and here we need to find the velocity of the each body at the time of meeting .

Solution

For the vertically thrown body, acceleration due to gravity shall be treated as negative and for the vertically thrown down body it is positive. For the second body its initial velocity is zero. Equating the equations of both the cases, we can find the specific time after which the stones are going to meet. Again by using that time value, we can find velocities of the bodies using the first equation of the motion as shown in the diagram below.

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