Question

A ball is thrown vertically upwards with the velocity of 25 m s. Calculate the time taken by the ball to reach the maximum height.
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Answer 1

Answer:

5sec

Explanation:

Initial velocity of the ball final velocity =0m/s

acceleration due to gravity (g)=−10m/s

2

(upward motion)

By using the formula,

v=u+at

O=25+(−10)×t

10t=25

t=2.5sec.

The time taken for upward journey is equal to the time taken for downward journey.

Therefore, total time taken by the ball =2.5+2.5=5sec.

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Answer 2

Correct Question:-

A ball is thrown vertically upwards with the velocity of 25 m/s. Calculate the time taken by the ball to reach the maximum height.

Given data:-

A ball is thrown vertically upwards with the velocity of 25 m/s.

—› Initial velocity ( u ) = 25 m/s

—› Final velocity ( v ) = 0

Solution:-

Let, acceleration due to gravity oppose the ball hence,

—› a = - g ........( 1 )

{g = acceleration due to gravity = 9.8 m/s²}

Now, we use scalar form for motion in one dimension ( linear motion )

—› v² = u² + 2as

{ This equation become }

—› v² - u² = 2as .........( 2 )

{ where, a = acceleration & s = displacement of ball }

Now, from eq. ( 2 )

—› v² - u² = 2as

{ from given & eq. ( 1 ) }

—› (0)² - (25)² = 2× ( - g ) × s

—› 0 - 625 = 2× ( - 9.8 ) × s

—› - 625 = - 19.6 × s

{ minus ( - ) sign cancel from both side }

—› 625 = 19.6 × s i.e.

—› s = 625 /19.6

—› s = 31.8877 m ( approx )

Now, to find time taken by ball in an upword direction we use formula :

—› v = u + at

—› v - u = at

{ from given & eq. ( 1 ) }

—› 0 - 25 = ( - g ) × t

—› - 25 = - 9.8 × t

{ minus ( - ) sign cancel from both side }

—› 25 = 9.8 × t i.e.

—› t = 25/9.8

—› t = 2.5510 sec ( approx )

Hence,the 2.5510 sec (approx) taken by the ball to reach the 31.8877 m (approx) height.