Question

A ball is thrown upwards with the velocity of 29.4 m/sec in the absemce of air resistence .find the time taken by the ball to reach the groun

Answer 1

the total time taken=2u/g
Here, u=29.4m/s
total T=29.4×2/g
=58.8/g s
g=9.8m/s²
so,. total t= 58.8/9.8=6s

Answer 2

Correct Question:-

A ball is thrown upwards with the velocity of 29.4 m/s in the absence of air resistence . find the time taken by the ball to reach the ground.

Given data:-

A ball is thrown vertically upwards with the velocity of 29.4 m/s.

—› Initial velocity ( u ) = 29.4 m/s

—› Final velocity ( v ) = 0

Solution:-

As we know upword acceleration is opposite to downword acceleration.

Let, a be the upword acceleration and g be the downword acceleration. 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 - (29.4)² = 2× ( - g ) × s

—› 0 - 864.36 = 2× ( - 9.8 ) × s

—› - 864.36 = - 19.6 × s

{ minus ( - ) sign cancel from both side }

—› 864.36 = 19.6 × s i.e.

—› s = 864.36/19.6

—› s = 44.1 m ( approx )

Now, to find time taken by ball when it thrown in an upword direction and return to ground, we use formula :

—› v = u + at

—› v - u = at

{ from given & eq. ( 1 ) }

—› 0 - 29.4 = ( - g ) × t

—› - 29.4 = - 9.8 × t

{ minus ( - ) sign cancel from both side }

—› 29.4 = 9.8 × t i.e.

—› t = 29.4/9.8

—› t = 3 sec ......( 3 )

{From eq. ( 3 )}

—› Total time take by ball

= upword direction + downword dirction

—› Total time take by ball = [ 3 + 3 ] sec

—› Total time take by ball = 6 sec

Hence,the time taken by the ball to reach the ground is 6 sec.