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# Motion Under Gravity – Steps, Impact & Importance

## Key Concepts:

•  Motion under gravity
•  Problems based on the motion under gravity
•  The graphical presentation of motion under gravity

### Introduction:

The acceleration due to gravity ’g’ remains constant near the surface of the earth. So, all the practical examples of objects executing vertical motion under gravity can be taken as one-dimensional motion with constant acceleration. We can study motion under gravity using the kinematic equations.

We will study the two cases of motion under gravity.

Case 1: An object is released downward

Case 2: An object is thrown upward

## Explanation:

### Case 1: An object released downward

We know that the three kinematic equations to use when an object executes uniformly accelerated motion are:

v = u + at

s = ut +𝟏/𝟐 at2

v2 – u2 = 2as

For an object falling under gravity, acceleration = a = g,

Height = s = h

so the equations become:

v = u + gt

h = ut +𝟏/𝟐 gt2

v2 – u2 = 2gh

Suppose the object is released from rest, then the initial velocity = u = 0.

v = gt

h = 𝟏/𝟐gt2

v2 = 2gh

Suppose the object is released from rest, then the initial velocity = u = 0.

v = gt

h = 𝟏/𝟐gt2

v2 = 2gh

If the total time to reach the ground = t = T

h = 𝟏/𝟐 gt2

Total time = T = √𝟐𝒈𝒉

Maximum speed with which it will reach the ground = vmaximum

v2 = 2gh ⟹ vmaximum = √𝟐𝒈𝒉

Problem 1: Two balls of masses 50gm and 100gm are dropped from the same height of 80m, find the time taken by the balls to reach the ground. (Ignore air resistance and take g = 10 m/s2).

Solution:

Since the equations of motion under gravity are independent of the mass of the object, the time taken by the ball to reach the ground will be the same and that is given by the equation:

T = √2h/g

=√2×80/10 = 4 sec

Thus, both the 50gm and 100gm balls will reach the ground after 4 seconds.

### Case 2: An object thrown upward

We know that the three kinematic equations to use when an object executes uniformly accelerated motion are:

v = u + at

s = ut + 𝟏/𝟐 at2

v2 = u + 2as

For an object moving against gravity, acceleration = a = – g,

Height = s = h

So, the equations become:

v = u – gt

h = ut – 𝟏/𝟐 gt2

v2 = u2 – 2gh

When a ball is thrown upwards with velocity u, it moves against gravity. Gradually its velocity decreases and after reaching a certain height its velocity becomes zero.

Initial velocity = u

At the maximum height attained h = H

Final velocity = v = 0

Time of ascent = ta

v = u + gt

0 = u – gta

Time of ascent = ta = u/g

Also, v2 – u2 = 2as

0 – u2 = –2gH

u = √2gH

Thus, time of ascent = ta = √2gH/g

Time of ascent = ta = √2H/g

When the ball falls, initial velocity = u = 0 and a = g

Time of descent = t = td

v = u + gtd

v = gtd

H = 1/2gtd2

td = √2H/g

Thus, we see that

Time of ascent = Time of descent

ta = td = √𝟐𝑯𝒈

Problem-2: A ball is thrown up with a speed of 40 m/s. Calculate the time of ascent and the maximum height it reaches. (Ignore air resistance and take g = 10 m/s2).

Solution:

Initial velocity = u = 40m/s

At the maximum height attained h = H

Final velocity = v = 0

Time of ascent = ta

v = u + gt

0 = u – gta

Time of ascent = ta = u/g = 40/10 = 4 s

Also, v2 – u2 = 2as

0 – u2 = –2gH

–402 = –2 × 10 × H

1600 = 20H

H = 80m, the ball will reach the maximum height of 80m.

### Graphical presentation of motion under gravity:

#### Case 1: Graphs for a body released downward:

Here, the graphs show the motion of a ball influenced by gravity alone. The ball is released from a height of 20 meters above the ground. The position is marked at 0.5 s intervals, the ball’s position, velocity, and acceleration are shown as a function of time.

Table 1: For a body released downwards

#### Case 2: Graphs for a body thrown upwards:

Here the graphs show the motion of a ball influenced by gravity alone. The ball is launched straight up from the ground with an initial speed of 20m/s. The position is marked at 0.5 s intervals, the ball’s position, velocity, and acceleration are shown as a function of time.

Table 2: For a body thrown upwards

Numerical 1: A ball of mass 200gm is thrown vertically up from the ground. It reaches a maximum height of 20m in 10s. Find the initial velocity of the ball.

1. 20m/s
2. 10 m/s
3. 15 m/s
4. 12 m/s

The initial velocity of the ball = u

Final velocity of the ball = v = 0 m/s

Acceleration of the ball = a = g = –10 m/s2 (negative sign because it is moving against gravity)

Maximum height attended = h = 20m

Using equation v2 – u2 = 2gh

0 – u2 = 2 × (–10) × 20

–u2 = –400

u = 20 m/s

Thus, the initial velocity of the ball = 20 m/s

Numerical 2: A ball is thrown vertically down. Find the time taken by the ball to reach the ground from a height of 500m.

1. 20s
2. 10s
3. 5s
4. 5s

When an object is just released from a certain height the initial velocity of the object = u = 0

The final velocity of the stone = v

Acceleration of the stone = a = g = 10 m/s2 (positive sign because it is moving in the direction of gravity)

Height = h = 500m

Using equation h = ut + ½ gt2

500 = 0 × t + ½ × 10 × t2

500 = 0 + 5t2

100 = t2

Time = t = 10 m/s

## Summary:

• The acceleration due to gravity ‘g’ remains constant near the surface of the earth. So, all the practical examples of objects executing vertical motion under gravity can be taken as one-dimensional motion with constant acceleration.
• For an object to fall under gravity:

Acceleration = a = g, Height = s = h, initial velocity =u = 0

v = gt, h = gt2, v = 2gh2

• If the total time to reach the ground = t = T

h = 1/2gt2

• For an object moving against gravity

Acceleration = a = -g, Height = s = h

so the equations become:

v= u – gt

h = ut – 1/2gt2

v2= u2-2gh

Time of ascent = Time of descent

ta = td = √2H/g, Total time = T = √2gh

Maximum speed with which it will reach the ground = vmaximum

v2 = 2gh = v maximum = √2gh

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