Question

A ball is thrown vertically upwards with the speed of 30 ms-1 . It returns back to the hands of the thrower after 6 s . Calculate the maximum height attained by the ball ( take g = 10 ms-2)

Answer 1

Given :-

A ball is thrown vertically upward with the speed of 30 m/s .

It returns back to the hands of the thrower after 6 seconds.

Required to find :-

• Maximum height attained by the ball ?

Equation used :-

v² - u² = 2as

Solution :-

Given data :-

A ball is thrown vertically upward with the speed of 30 m/s .

It returns back to the hands of the thrower after 6 seconds

we need to find the maximum height attained by the ball

So,

From the given information we can conclude that ;

• Initial velocity of the ball ( u ) = 30 m/s

• Final velocity of the ball ( v ) = 0 m/s

Since, the ball is moving against the acceleration due to gravity . So, we need to take the value of g in terms of negative .

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

Using the equation of motion ;

i.e. v² - u² = 2as

☞ ( 0 )² - ( 30 )² = 2 x - 10 x s

☞ 0 - 900 = - 20 x s

☞ - 900 = - 20s

☞ Taking - ( minus ) Common

☞ we get ;

☞ - ( 900 ) = - ( 20s )

☞ - ( minus ) get's cancelled on both sides

☞ 900 = 20s

☞ 20s = 900

☞ s = 900/20

☞ s = 90/2

☞ s = 45 meters

Hence,

Displacement ( s ) = 45 meters

Alternative method :-

Using the 2nd equation of motion ;

i.e s = ut + ½ at²

Since,

The time taken by the ball to reach the maximum height = Total time /2

=> 6/2

=> 3 seconds

This implies,

✐ s = 30 x 3 + ½ x - 10 x 3 x 3

✐ s = 90 + ½ x - 30 x 3

✐ s = 90 + ½ x - 90

✐ s = 90 + ( - 45 )

✐ s = 90 - 45

✐ s = 45 meters

Therefore,

Maximum height attained by the ball is 45 meters .

Answer 2

☣️$\huge\rm\blue{GIVEN}$

• ➡️$\rm\blue{Initial\:Velocity=10m/s^-1}$

• ↪️$\rm\red{Final\:Velocity=0m/s^-1}$

• ↪️$\rm\red{Time\:Taken=6s}$

• As the Acceleration is acting against the gravity,So we will take it as a retardation.

• ↪️$\rm\green{Acceleration= -10m/s^-2}$

☣️$\huge\rm\blue{To\:Find}$

• Maximum eight attained by the ball.

☣️$\huge\tt\blue{FORMULAE\:USED}$

${\boxed{\tt{\red{S=ut+\dfrac{1}{2}\times{a}\times{(t)}^2}}}}$

Where,

S= Distance

u= initial Velocity

t= Time taken

a= Acceleration

Since, Time taken to reach the maximum height =6/2= 3s.

$\implies\tt\blue{S=ut+\dfrac{1}{2}\times{g}\times{(t)}^2}$

$\implies\tt\blue{S=30\times{3}+\dfrac{1}{\cancel{2}}\times{\cancel{-10}}^-5\times{9}}$

$\implies\tt\blue{S=90-45}$

$\implies\tt\blue{s=45m}$

Hence, The height attained by baal is 45m

$\large\mathbb\blue{EXTRA\:INFORMATION}$

• When the object falls freely we take initial velocity as zero.

• When object is thrown upwards we taken Final Velocity as zero.