Question

A ball is dropped from rest on a cliff a)what is the velocity of the ball 5 seconds later?
b)How far does it travel during this time?

Answer 1

$\bigstar\large\boxed{\mathtt{\green{Given:}}}$

• Initial velocity of the ball = 0 (As the ball is dropped from rest )

• Acceleration due to gravity = 9.8 m/s

• Time taken = 5 sec

$\bigstar\large\boxed{\mathtt{\blue{To find :}}}$

• Velocity of the ball after 5 seconds
• Distance covered by the ball in 5 sec

$\bigstar\large\boxed{\mathtt{\pink{How\: to \:solve :}}}$

• As the acceleration due to gravity is constant in both the cases we can use the equations of motion in order to solve this question
• Part I :- As we already know the values of u,g,ant t we can use the first equation of motion in order to find v
• Part II :- As we know the values of u, t , g we can use the second equation of motion in order to find s

$\bigstar\large\boxed{\mathtt{\orange{Solution :}}}$

Part I :-

By using the first equation of motion,

$\implies\boxed{\mathtt{ v = u +gt }}$

here,

• v = final velocity
• u = initial velocity
• g = acceleration due to gravity
• t = time

$\implies\mathtt{ v = 0 + 9.8 \times 5 }$

$\implies\mathtt{ \red{v = 49 \dfrac{m}{s} }}$

The velocity of the ball after 5 seconds will be 49 m/s

------------------------------

Part II :-

As the acceleration due to gravity remains constant throughout the motion we can easily solve this question by using the second equation of motion ,

$\implies\boxed {\mathtt{ s = ut +\dfrac{1}{2}at^{2} }}$

here ,

• s = distance
• u = initial velocity
• t = time taken
• a = acceleration

$\implies\mathtt{ s = 0+\dfrac{1}{2}\times 9.8\times 25 }$

$\implies\mathtt{ s = 4.9\times 25 }$

$\implies\mathtt{\red{ s = 122.5 m }}$

The distance traveled by the ball in 5 sec = 122.5 m

$\bigstar\large\boxed{\mathtt{Additional \: formulas :}}$

Third equation of motion

$\implies\mathtt{ v^{2}=u^{2} + 2as }$

Distance traveled by the object in nth sec

$\implies\mathtt{ s = u + \dfrac{a}{2}(2n-1)}$

Answer 2

Concept:

Equations of motion,

1) v = u + at

2)s = ut + ½at²

3) v² = u² + 2as

Given:

Initial velocity of the ball, u = 0 ms⁻¹

Acceleration due to gravity, g = 9.8 ms⁻²

Time interval, t = 5 s

Find:

a) Velocity of the ball after 5 seconds.

b) Distance covered by it in 5 seconds.

Solution:

a) Using the first equation of motion,

v = u + at

where v = Final velocity (velocity after 5 seconds)

u = Initial velocity = 0 ms⁻¹

a = g = Acceleration due to gravity [as body is falling under gravity] = 9.8 ms⁻²

t = Time interval = 5 s

Putting all value in equation, we get

v = 0 + (9.8)(5)

v = 49 ms⁻¹

Hence, the velocity of ball after 5 seconds is 49 ms⁻¹.

b) Using the second equation of motion,

s = ut + ½at²

where s = distance covered in 5 seconds

v = Final velocity (velocity after 5 seconds)

u = Initial velocity = 0 ms⁻¹

a = g = Acceleration due to gravity = 9.8 ms⁻²

t = Time interval = 5 s

Putting all values in above equation, we get

s = (0)(5) + ½(9.8)(5)²

s = 0 + ½(9.8)(25)

s = ½(9.8)(25)

s = ½(245)

s = 122.5 m

Hence, the distance covered by the ball in 5 seconds = s = 122.5 m

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