Question

a ball thrown vertically upward
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Answer 1

Answer- The above question is from the chapter 'Kinematics'.

Some important terms and formulae:-

1. Velocity- It is the displacement per unit time.

S.I. Unit of Velocity- m/s

It is a vector quantity as it possesses magnitude and direction.

2. Acceleration- It is the rate of change of velocity.

S.I. Unit of Acceleration- m/s²

It is also a vector quantity.

Negative acceleration is called retardation.

3. Distance- It is the path length transversed by an object.

S.I. Unit of Distance- m

It is a scalar quantity.

4. Displacement- It is the shortest distance between the initial and final point.

S.I. Unit of Displacement- m

It is a vector quantity.

5. Equations for uniformly accelerated motion-

Let u = Initial velocity of a particle

v = Final velocity of a particle

t = Time taken

s = Distance travelled in the given time

a = Acceleration

1) v = u + at

2) s = $\frac{1}{2}$ at² + ut

3) v² - u² = 2as

Concept used: 1) Velocity of a body thrown upwards at the highest point is 0 m/s.

2) Three equations of motion

Given question: 1) A body thrown vertically upwards reaches the highest point of its path in 2 s. Find the initial velocity with which the ball is thrown. Also, find the height of this highest point. (Acceleration due to gravity = 10 m/s²)

2) A stone is thrown down vertically from the top of a tower with initial velocity of 10 m/s. It strikes the ground after 2 seconds. Calculate the velocity with which the stone strikes the ground and the height of the tower.

Answer: 1) Let the initial velocity of the ball be = u.

Final velocity (v) = 0 m/s

Time taken (t) = 2 s

Acceleration due to gravity (a) = -10 m/s²

Using 1st equation of motion, v = u + at, we get,

0 = u + (-10) × 2

u = 20 m/s

Let the height of the highest point be h.

Using third equation of motion, v² - u² = 2ah, we get,

0² - 20² = 2 × -10 × h

h = -400 ÷ -20

h = 20 m

∴ Initial velocity of the ball = 20 m/s.

Height of the highest point = 20 m.

2) Initial velocity of the stone (u) = 10 m/s

Let the final velocity of the stone be v.

Time taken (t) = 2 s

Acceleration due to gravity (a) = 10 m/s²

Using 1st equation of motion, v = u + at, we get,

v = 10 + 10 × 2

v = 10 + 20

v = 30 m/s

Let the height of the tower be h.

Using third equation of motion, v² - u² = 2ah, we get,

30² - 10² = 2 × 10 × h

900 - 100 = 20h

h = 800 ÷ 20

h = 40 m

∴ Final velocity of the stone = 30 m/s.

Height of the tower = 40 m.

Answer 2

Answer- The above question is from the chapter 'Kinematics'.

Some important terms and formulae:-

1. Velocity- It is the displacement per unit time.

S.I. Unit of Velocity- m/s

It is a vector quantity as it possesses magnitude and direction.

2. Acceleration- It is the rate of change of velocity.

S.I. Unit of Acceleration- m/s²

It is also a vector quantity.

Negative acceleration is called retardation.

3. Distance- It is the path length transversed by an object.

S.I. Unit of Distance- m

It is a scalar quantity.

4. Displacement- It is the shortest distance between the initial and final point.

S.I. Unit of Displacement- m

It is a vector quantity.

5. Equations for uniformly accelerated motion-

Let u = Initial velocity of a particle

v = Final velocity of a particle

t = Time taken

s = Distance travelled in the given time

a = Acceleration

1) v = u + at

2) s = \frac{1}{2}

2

1

at² + ut

3) v² - u² = 2as

Concept used: 1) Velocity of a body thrown upwards at the highest point is 0 m/s.

2) Three equations of motion

Given question: 1) A body thrown vertically upwards reaches the highest point of its path in 2 s. Find the initial velocity with which the ball is thrown. Also, find the height of this highest point. (Acceleration due to gravity = 10 m/s²)

2) A stone is thrown down vertically from the top of a tower with initial velocity of 10 m/s. It strikes the ground after 2 seconds. Calculate the velocity with which the stone strikes the ground and the height of the tower.

Answer: 1) Let the initial velocity of the ball be = u.

Final velocity (v) = 0 m/s

Time taken (t) = 2 s

Acceleration due to gravity (a) = -10 m/s²

Using 1st equation of motion, v = u + at, we get,

0 = u + (-10) × 2

u = 20 m/s

Let the height of the highest point be h.

Using third equation of motion, v² - u² = 2ah, we get,

0² - 20² = 2 × -10 × h

h = -400 ÷ -20

h = 20 m

∴ Initial velocity of the ball = 20 m/s.

Height of the highest point = 20 m.

2) Initial velocity of the stone (u) = 10 m/s

Let the final velocity of the stone be v.

Time taken (t) = 2 s

Acceleration due to gravity (a) = 10 m/s²

Using 1st equation of motion, v = u + at, we get,

v = 10 + 10 × 2

v = 10 + 20

v = 30 m/s

Let the height of the tower be h.

Using third equation of motion, v² - u² = 2ah, we get,

30² - 10² = 2 × 10 × h

900 - 100 = 20h

h = 800 ÷ 20

h = 40 m

∴ Final velocity of the stone = 30 m/s.

Height of the tower = 40 m.