Question

A ball is dropped from a height of 125m above the ground. Calculate 1. Distance travelled by ball in 5th second of free fall. 2. Speed of the ball when it hits the ground. 3. Time for which ball remains in air.

Answer 1

✦ Given ✦

A ball is dropped from a height of 125 m above the ground.

✦ To Find ✦

• Distance travelled by ball in 5th second of free fall.
• Speed of the ball when it hits the ground.
• Time for which ball remains in air.

✦ Solution ✦

We know,

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

◘ Initial velocity, u = 0 m/s

◘ Acceleration, a = g = 10 m/s²

◘ Time, t = 5 s

s = 0(5) + 1/2 10(5)²

⇒ s = 5(5)²

⇒ s = 125 m

___________________

As from above, the ball hits the ground in the 5th second (it covered 125 m in the 5th second which was the height from which the ball was dropped off).

We know,

$\boxed{\sf{v=u+at}}$

◘ u = 0 m/s

◘ a = 10 m/s²

◘ t = 5 s

⇒ v = 0 + 10(5)

⇒ v = 50 m/s

___________________

Now, as the ball took 5 secs to touch the ground, then the time for which the ball remained in the air = 5 - 1

= 4 secs

Answer 2

Given :-

A ball is dropped from a height of 125 meters above the ground .

Required to find :-

• Distance travelled by the ball in the 5th second ?

• Speed of the ball when it hits the ground ?

• Time for which ball remains in air ?

Equation used :-

v = u + at

s = ut + ½ at²

v² - u² = 2as

Solution :-

Given data :-

A ball is dropped from a height of 125 meters above the ground .

we need to find the ;

Distance travelled by the ball in the 5th second ?

Speed of the ball when it hits the ground ?

Time at which the ball remains in air ?

So,

From the given information we can conclude that ;

• Initial velocity of the ball ( u ) = 0 m/s

• Displacement ( s ) = 125 meters

Since, the ball is falling freely the acceleration due to gravity must be taken in positive .

So,

• Acceleration due to gravity ( g ) = 10 m/s²

Now,

Using the equation of motion let's find the time taken by the ball

The equation which we are going to use is s = it + ½ at²

s = 0 x t + ½ x 10 x t x t

125 = 0 + ½ x 10 x t²

125 = 0 + 5 x t²

125 = 0 + 5t²

125 = 5t²

5t² - 125 = 0

Taking 5 common

5 ( t² - 25 ) = 0

t² - 25 = 0/5

t² - 25 = 0

( t )² - ( 5 )² = 0

Since, we know that

a² - b² = ( a + b ) ( a - b )

( t + 5 ) ( t - 5 ) = 0

This implies ;

t + 5 = 0

t = - 5

t - 5 = 0

t = 5

Since, time can't be in negative .

So,

• Time taken by the ball to fall = 5 seconds

Now,

Let's find the speed of the ball when it hits the ground ;

Using the equation of motion ;

i.e. v² - u² = 2as

v² - ( 0 )² = 2 x 10 x 125

v² - 0 = 2 x 10 x 125

v² = 20 x 125

v² = 2500

v = √2500

v = 50

Hence,

• Final velocity of the ball ( v ) = 50 m/s

Similarly,

It is also given that we need to find the displacement at the 5th seconds .

But, according to above calculations we can say that the ball had displaced 125 meters in 5 seconds .

Let's get it numerically too ;

Using the 2nd equation of motion ;

s = ut + ½ at²

s = 0 x 5 + ½ x 10 x 5 x 5

s = 0 + ½ x 10 x 25

s = 0 + 5 x 25

s = 0 + 125

s = 125 meters

Hence, the displacement of the ball in 5 seconds is 125 meters

Therefore,

Distance travelled by the ball in the 5 seconds is 125 meters

Speed of the ball when it hits the ground is 50 m/s

Time at which the ball remains in the air = 5 - 1 = 4 seconds