We would like to solve the first problem basing on one dimensional body. The problem is about a body thrown initially with zero velocity. After two seconds another body is dropped with a certain velocity. We need to know the velocity of that body so that both the bodies reach the ground simultaneously.

**Problem One**

**Solution**

The first body is initially has zero velocity and using the second equation of motion, we can find the time taken by the body to reach the bottom of the bridge and strike the water. It is given that the second stone is thrown down with a initial velocity after two seconds. So the second body takes only three seconds to reach the ground. Both the stones reach the bottom of the bridge so that the distance travelled by both of them is same.

By substituting them in the same second equation of motion, we can find the initial velocity with which the second stone is thrown from the same place as shown in the diagram below.

**Problem Two**

This problem of one dimensional motion along Y axis is also of the similar model of the previous problem. But in this problem, both the stones are falling freely so that have no initial velocity. The second stone is thrown few seconds later and we would like to know the distance of separation between the stones has to be a certain value after a specified time and we would like to find it out.

**Solution**

As per second equation of motion, we have relation between displacement, initial velocity, acceleration and time as shown in the diagram below. The second equation is different from the first as there is a time gap between both the stones.

It is given that after a certain time, we need to have a certain difference of separation. Thus by subtracting them we can solve the problem as shown in the diagram below.

**Problem Three**

This problem is also about two stones falling from a certain height and the second stone is having a falling distance certain separation 10 meter when compared with the first one.

It is given in the problem that both of them reach the bottom of the tower at the same time and the second stone is actually thrown one second later than the first one.

**Solution**

For a freely falling body, by using the second equation of motion, we can find the height travelled by the body before it had reached the bottom. For the second stone also, we shall apply the same equation, but the time is one second less and height is 10 meter less.

By substituting the value of the height from the first equation in the second equation, we can solve the problem as shown in the diagram below.We can find both the time as well as height of the tower.

**Problem Four**

The problem is about the distance travelled by the body in the particular second. We have forth equation of motion in one dimensional motion. It is given that in the given second a certain distance of seven meter and in the previous second five meter has fallen. We would like to measure the velocity of the body with which it strikes the ground.

**Solution**

By solving them using forth equation of motion as shown in the diagram below after dividing both the cases, we can get the value of the n th second. Further we can find the final velocity of the body using the first equation of motion. The solution is shown in the diagram below.

**Problem Five**

It is given in the problem that one stone is dropped from a certain height 180 meter. It has no initial velocity. Another stone is dropped from a certain point below the first height and it means, it is travelling less distance than the first one. Both of them reached the ground at the same time but the second one got started initially late by two seconds.

**Solution**

By using the second equation of the motion, we can write two equations as shown in the diagram below. By solving both the equations, we can solve the problem as shown in the diagram below.

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