Sunday, July 5, 2015

Acceleration due to gravity and One Dimensional Motion Equations

Acceleration due to gravity

The Earth is a huge massive body. According to Newton law of gravitation, the force of attraction between the two bodies is directly proportional to the product of the masses and inversely proportional to the square of the distance of separation. Here the two masses involved are the mass of the body and the mass of the Earth. The distance between the bodies is generally radius of the Earth, as the body is on the surface of the earth or near to the surface of the earth. As the mass of the earth is comparatively very high when compared with the body, the gravitational force of attraction will be always towards the Centre of the Earth. This gravitational force provides acceleration to all the bodies on the Earth and the corresponding acceleration is called acceleration due to gravity. The numerical value of acceleration due to gravity on the Earth is 9.8 m/s Squire. It may vary slightly from place to place but in general it can be taken as a constant.

This acceleration on the surface of the earth is uniform and constant. Hence when the body is coming towards the Earth, because of acceleration due to gravity its velocity will be keep on increasing until it strikes the ground. If you throw the body against the acceleration due to gravity, into the sky, it’s velocity will be keep on decreasing and finally it stops at a certain height called maximum height. Time taken to reach that maximum height is called time of ascent and time taken to reach the ground from a certain height is called time of descent.

We have four different equations of the motion to express the translatory motion of a body. In all that equations where ever acceleration is there, we can substitute acceleration due to gravity and we can rewrite the corresponding equations of motion due to gravity. Here there are practically two possibilities. The body may be coming towards the Earth or the body may be going away from the Earth. The acceleration due to gravity on the bodies that are coming towards the earth is generally treated as positive and vice versa.

We can write the equations of motion and the corresponding values as shown below.

Using this equations we can find the time taken by the body to reach a particular height against the gravity and it is called time of ascent. If the body is falling from the same height to reach the ground to take the same time and that is called time of dissent. The sum of the time of ascent and time of dissent is called time of flight.

A body is said to be a freely falling body if it is falling from a certain height with zero initial velocity. In that case, using the equations of motion we can calculate the final velocity of the body after covering a certain height h.

It can be also proved that if your body is thrown up with the velocity from the ground, after reaching the maximum height it will come back to the ground with the same velocity but in the reverse direction.

In general we have ignored the impact of the air resistance while measuring the time of ascent and time of descent. If we consider the air resistance, then the time of ascent is going to be little bit different from time of dissent. The direction of the force due to the air resistance is always in opposition to the relative motion.

So the effective acceleration in this case is going to be more than acceleration due to  gravity and hence the time of ascent is less than that of the case when the air resistance is ignored. When we are ignoring air resistance we are imagining that the environment is vacuum for the calculation purpose. Though it is not practically vacuum the given equations will be approximately valid in almost all the real-time situations.

By taking their resistance into consideration if we try to calculate the time of dissent, being their resistances against the relative motion, it is in the upward direction but the gravity is in the downward direction. And hence the effective acceleration will be less and hence time of dissent will be more and more than the case of vacuum.

It can be also be noticed that time of decent is more than the time of recent when their resistances taken into consideration.

Problem and solution

Two balls are dropped from different heights. One ball is drop two seconds after the other but both the strikes the ground at the same time which is five seconds after the first ball. Find the heights from where these two balls dropped?

As the first stone is taking five seconds to reach the ground and the second stone is two seconds late, it shall be only taking three seconds to reach the ground. The corresponding equations for the heights of the bodies can be written basing on the equation of motion as shown below. Simply the difference between the heights of the two equations we can calculate it as shown below.

Problem and solution

A body balls freely from a height of 125 m. After two seconds gravity ceases to act. Find the time taken by it to reach the ground if acceleration due to gravity is considered as 10 metre per second Squire.

The body is a freely falling body means its initial velocity equal to 0. For the first two seconds acceleration due to gravity is acting and hence it falls due to the gravity and it covers a distance of 20 m. During this process the body will acquire some velocity and it can be calibrated using the equations of motion as shown below. It is found that its value equal to 20 meter per second. Being it is drop from 125 m and 20 m is covered in the first two seconds, it has to further cover a distance of 105 m per reach the ground.

As acceleration due to gravity is no more acting the velocity of the body will remain constant. Does the body will continue to move with the same velocity of 20 m/s for the remaining time. We can calculate the time taken to cover the 105  m is in the formula that the displacement equal to velocity multiplied by time. By adding this time of the two seconds with can get the total time of the journey.

Problem and solution

A parachutist drops freely from an aeroplane for 10 seconds before the parachute opens out. He’s velocity when he reach the ground is 8 m/s. Once if the parachute is open he has a standard retardation of two meter per second Squire. Find the height at which he gets out of the aeroplane.

Before the parachute opens he falls down due to the gravitational effect and he’s not having a initial velocity in the vertically downward direction. Hence we can find that is going to cover a distance of 490 m in the first 10 seconds. In this 10 seconds will also acquire some velocity due to acceleration due to gravity and it can be found that the velocity is 98 m/s.

He covers the further distance with the standard retardation of two meter per second Squire and we can calculate the distance that he covers with a initial velocity of 98 m/s and a final velocity of 8 m/s and with an acceleration of two meter per second Squire using the equation of motion. By substituting the value is we can get the total height as 2875 m.

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