The motion under Gravity Notes. Gravity is one of the four fundamental forces of nature, and its effects can be seen in everyday life. From the Earth orbiting the Sun to the movement of the planets around their stars, gravity is the mysterious force that keeps the universe in motion. In this series of notes, we’ll explore the basics of gravity and its effects on motion. From Newton’s laws of motion to the motion of planets, we’ll investigate the mysteries of gravity and learn how it affects the motion of objects.Lets take drive into the short and crisp notes on Motion under Gravity Notes

**Motion under Gravity**

The motion under gravity refers to the movement of an object that is subject to the force of gravity. Here are some important concepts to keep in mind when studying motion under gravity.

**Acceleration due to Gravity**

Acceleration due to gravity is the acceleration experienced by an object when it falls towards the earth. It is denoted by the letter ‘g’ and has a value of approximately 9.8 m/s^2. The value of ‘g’ is constant at all points on the surface of the earth, and it is the same for all objects regardless of their mass.

**Free Fall**

Free fall refers to the motion of an object that is falling under the influence of gravity alone. In free fall, an object’s velocity increases as it falls towards the ground. When an object is in free fall, the only force acting on it is the force of gravity, and it accelerates at a constant rate of ‘g’ towards the ground.

Equations of Motion under Gravity

The following equations can be used to describe the motion of an object under gravity:

v = u + gt

s = ut + 0.5gt^2

v^2 = u^2 + 2gs

where:

v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time taken

s = distance travelled

These equations can be used to calculate the displacement, velocity, and acceleration of an object under gravity. The first equation gives the final velocity of the object, the second equation gives the displacement of the object, and the third equation gives the final velocity of the object squared.

**Projectiles**

A projectile is any object that is thrown into the air and then moves under the influence of gravity alone. The motion of a projectile can be described by the following equations:

Range = u^2 sin(2θ)/g

Time of flight = 2u sinθ/g

Maximum height = u^2 sin^2θ/2g

where:

θ = angle of projection

u = initial velocity

g = acceleration due to gravity

These equations can be used to calculate the range, time of flight, and maximum height of a projectile. The range is the horizontal distance travelled by the projectile, the time of flight is the time taken by the projectile to reach the ground, and the maximum height is the highest point reached by the projectile.

I hope these notes are more detailed and helpful for you. If you have any further questions or need more information, feel free to ask.

**Basic Numericals on Motion under Gravity **

Sure, here are some basic numerical problems on motion under gravity that involve solving for various parameters:

A ball is thrown vertically upward with an initial velocity of 20 m/s. How high does it rise before it starts falling back down?

Solution:

Using the formula for vertical displacement, we have:

v_f^2 = v_i^2 + 2gh

where v_f is the final velocity (0 m/s), v_i is the initial velocity (20 m/s), g is the acceleration due to gravity (-9.81 m/s^2), and h is the maximum height reached by the ball.

Solving for h, we get:

h = (v_f^2 – v_i^2)/(2g)

h = (0 – 20^2)/(2*(-9.81))

h = 20.2 meters

A stone is dropped from a height of 100 meters. How long does it take to reach the ground?

Solution:

Using the formula for vertical displacement, we have:

h = v_it + (1/2)gt^2

where v_i is the initial velocity (0 m/s), g is the acceleration due to gravity (-9.81 m/s^2), and h is the height from which the stone is dropped.

Solving for t, we get:

t = sqrt((2h)/g)

t = sqrt((2100)/9.81)

t = 4.52 seconds

A ball is thrown horizontally from the top of a cliff with a speed of 10 m/s. How far from the base of the cliff will the ball hit the ground if the cliff is 50 meters high?

Solution:

The horizontal motion of the ball is independent of its vertical motion, so we can use the formula for horizontal displacement, which is:

d = v_xt

where d is the horizontal distance traveled, v_x is the horizontal velocity (10 m/s), and t is the time taken for the ball to hit the ground.

Using the formula for vertical displacement, we can find the time taken for the ball to hit the ground:

h = (1/2)gt^2

where h is the height of the cliff (50 meters).

Solving for t, we get:

t = sqrt((2h)/g)

t = sqrt((2*50)/9.81)

t = 3.19 seconds

Now, using the formula for horizontal displacement, we have:

d = v_xt

d = 10 m/s * 3.19 seconds

d = 31.9 meters

Therefore, the ball will hit the ground 31.9 meters from the base of the cliff.

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