Question

A boy throws a ball up into the air. The initial velocity of the ball was 6 m/s. Taking g =10 ms- 1 and assuming there was no air resistance, find. (a) The maximum height the ball reaches, (b) The velocity of the ball when it reaches the boy's hand, (c) The time the ball is up in the air[Take the upwards motion to be negative] A. -1.8 m, 6 m/s, 1.2 s B. 1.8 m, 6 m/s, 0.6 s C.-1.8 m, 6 m/s, 0.6 s D. 1.8 m, 6 m/s, 1.2 s

Answer 1

Solution :-

Ball moving in upward direction

• Initial velocity (u) = 6 m/s

• Acceleration due to gravity (g) = -10 m/s ( as the ball is moving against gravity )

• Final velocity = 0 m/s ( velocity at the highest point is 0 )

A the ball moves uniformly under the influence of gravity we can use equations of motion in order to solve this question

(a)

By using the third equation of motion ,

➽ v² = u² + 2gh

➽ 0 =  u² + 2gh

➽ u² = - 2gh

➽ h = - u² / 2g

➽ h = - 36 / - 20

➽ h = 1.8 m

The maximum height the ball reaches = 1.8 m

(c)

By using the first kinematic equation ,

➽ v = u + at

➽ 0 = u + at

➽ u = - at

➽ 6 = - x - 10 x t

➽ t = 6 / 10

➽ t = 0.6 s

The time the ball is up in the air = 0.6 s

Ball moving in downward direction

• Initial velocity ( u ' ) = 0 m /s  ( velocity at the highest point is 0 )

• Acceleration due to gravity (g) = 10 m/s

• Time of descent = time of ascent = 0.6 s

(b)

By using the first kinematic equation ,

➽ v' = u ' + gt

➽ v ' = gt

➽ v ' = 10 x 0.6

➽ v ' = 6 m /s

The velocity of the ball when it reaches the boy's hand is 6 m/s

Answer 2

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: Question :- \: \star}}}}}$

A boy throws a ball up into the air. The initial velocity of the ball was 6 m/s. Taking g =10 ms- 1 and assuming there was no air resistance, find. (a) The maximum height the ball reaches, (b) The velocity of the ball when it reaches the boy's hand, (c) The time the ball is up in the air.

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

$\Large{\underline{\underline{\bf{GiVen:-}}}}$

❥ Initial velocity (u) = 6 m/s

❥ Acceleration due to gravity (g) = -10 m/s ( as the ball is moving against gravity )

❥ Final velocity = 0 m/s ( velocity at the highest point is 0 )

$\Large{\underline{\underline{\bf{Find:-}}}}$

(a) The maximum height the ball reaches,

(b) The velocity of the ball when it reaches the boy's hand,

(c) The time the ball is up in the air.

$\Large{\underline{\underline{\bf{SoLuTion:-}}}}$

Here we go !

❥ Initial velocity = 6 m/s

❥ Acceleration due to gravity (g) = -10 m/s

❥ Final velocity = 0 m/s

We can solve it by using the equation of motion :-

Case 1 :-

Applying 3 eqn . of motion

v² = u² + 2gh

↠0 =  u² + 2gh

↠ u² = - 2gh

↠ h = - u² / 2g

↠ h = - 36 / - 20

↠ h = 1.8 m

Therefore :-

A ) What is Maximum height the ball reaches ?

❥The maximum height the ball reaches = 1.8 m .

Case 2 :-

Initial velocity ( u ' ) = 0 m /s  ( velocity at the highest point is 0 )

Acceleration due to gravity (g) = 10 m/s

Time of descent = time of ascent = 0.6 s

Applying first kinematic equation ,

v' = u ' + gt

↠ v ' = gt

↠ v ' = 10 x 0.6

↠ v ' = 6 m /s

B) What is the velocity of the ball when it reaches the boy's hand ?

❥ The velocity of the ball when it reaches the boy's hand is 6 m/s.

Case 3 :-

Applying First equation of motion

v = u + at

↠ 0 = u + at

↠ u = - at

↠ 6 = - x - 10 x t

↠ t = 6 / 10

↠ t = 0.6 s

C ) what's the time the ball is up in the air ?

❥ The time the ball is up in the air = 0.6 s

Additional Information :-

$\\\\1) \: v = u + at \\ where \: \\ v \: = final \: velocity \\ u \: = initial \: velocity \\ a \: = acceleration$

$2) \: {v}^{2} = {u}^{2} + 2aS ...(i)\\\: \: or \: \: {v}^{2} = {u}^{2} + 2gh...(ii)\\ \: \\ if \: acceleration \: is \: given \: we \: use \: equation \: (i) \\ if \: acceleration \: due \: to \: gravity \: is \: given \: we \: use \: equation \: (ii)$

$3) \: S = ut + \frac{1}{2} a {t}^{2} \\ where \: S \: = distance \\ u \: = initial \: velocity \\ a \: = acceleration \\ t \: = time$

Explanation:

Derivation of Equation by Algebra Method:

(1)

We know that,

acceleration= rate of change of velocity

$a = \frac{v - u}{t}$

Cross multiplying, we get:

$at = v - u$

$v = u + at$

Hence Proved

(2)

We already have v = u + at.

We can write above equation as below

$t = \frac{v - u}{a}$

$\tt \: distance \: = \: average \: velocity \times time$

and

$\tt \: average \: velocity \: = \frac{initial \: velocity+ final \: velocity}{2}$

$\tt \: average \: velocity = \frac{v + u}{2}$

Therefore,

$\\\tt\: distance \: = \frac{v + u}{2} \times \frac{v - u}{a}$

$\tt \: S = \frac{{v}^{2} - {u}^{2} }{2a}$

Cross multiplying, we get:

$\\\tt \: {v}^{2} - {u}^{2} = 2aS$

If acceleration is not given and acceleration due to gravity is given , then instead of a we substitute the value of g (i.e. 9.8 m/s^2 ) and the value distance is substituted as h.

Hence Proved.

(3)

We know that,

$\\\tt \: distance = average \: velocity \: \times time$

$\tt \: average \: velocity = \frac{v + u}{2}$

$\sf \: S = (\frac{v + u}{2} ) \times t$

We know that,

Substituting the value of v in the above equation,

$\\\tt \: v = u + at$

$\sf \: S = ( \frac{u + at + u}{2} ) \times t$

$\sf \: S = (\frac{2u + at}{2}) \times t$

We can write above equation as below,

$\\\sf \: S =( u + \frac{1}{2} at)t$

Multiplying t , we get:

$\\\sf \: S = ut + \frac{1}{2} a {t}^{2}$

Hence Proved.

Special Cases:

If initial velocity is zero(i.e. body is at rest).

$\\\\\sf\:v\:=\:at$

$\sf\:v^2\:=\:2as\\or\\v^2\:=\:2gh$

$\sf\:S\:=\:\frac{1}{2} a t^2$