Question

A ball is dropped from a height of 125 m above the ground. the distance travelled by ball during the last second of its motion is

Answer 1

Answer:

In the first second, the stone travels S:

S = ut + 1/2 gt^2

S = 0t + 1/2 (10) 1^2

S = 5 meters.

Distance at the end of the first second is 5 meters.

The total number of seconds of fall is t:

S = ut + 1/2 gt^2

125 = 0t + 1/2 (10) t^2

25 = t^2

t = 5 seconds time of fall.

The distance traveled in the last second is the distance traveled by the end of the 5th second (125 meters) minus the distance traveled by the end of the 4th second.

4th second distance:

S = ut + 1/2 gt^2

S = 0t + 1/2 (10) 4^2

S = 5(16) = 80 meters

Distance at the end of 4 seconds is 80 meters.

Distance a...

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Answer 2

Answer:

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

Explanation: Please mark me as the brainlest