a ball of mass 100g is projected vertically upwards from ground with vel. 49m/s.

at the same time another identical ball is dropped from a height of 98m to fall freely along the same path as that of the first.the two balls collide and stick together and fall to ground.find time of flight of masses?

 

ans:6.53s

see when the 2 balls collide then due to their combined momentum, they do not stop but they attain a velocity that can be found out by
applying law of conservation of momentum i e

momentum bfore collision=momentum after collision
m.49+m.0=2m.v

therefore
v=49/2

it is in upward direction as it is + ve

with this velocity the body(combined mass) continues to move in upward direction till the time

0=49/2-9.8t

calculate t
and put it in in formula for s=ut+1/2at^2
u will get the distance now add this distance to the distance of collision from ground
and use the formula

t=root(2h/g)
add all times and u will get time of flight

hope u get it !!

u can ask if u did'nt

after paining my hands so much don't u think i deserve a rate...

First calculate highest point reached by first ball

v^2 - u^2 = 2as

- 49^2 = 2 .-9.8 .s

122.5 m = s

 

1/2 g t^2 = 122.5 - ut + 1/2 g t^2 

t = 2.5 secs

 

v = u + at

so v1 = -24,  v2 = 25

m1v1 + m2v2 = 2mV

0.1*-24 + 0.1*25 = 2* 0.1*V

V = 0.5 ( of combined mass)

 

now in 2.5 secs the dist covered was

1/2 a t^2... for 2nd ball

1/2*10*(2.5)^2

31.25 m

 

So remaining distance from ground

= 98 - 31.25 m

= 66.75 m

 

66.75 = 0.5t + 1/2*10*t^2

5t^2 + 0.5t - 66.75 = 0

t =  -0.5 + sqrt(0.25 + 1335)  / 10

t = 3.6 seconds

This is the time taken for the combined mass to fall to the ground.

now u need to add the time taken by the balls to meet

around 2.5 ..  i used approximations with g

so t = 3.6 + 2.5 = 6.1 sec in this case

here another sol..n u get the ans as 6.53 exact.

eqn fr motion of 1st ball:

s₁=49t - (1/2 gt²);

fr 2nd ball:

s₂=98 - (1/2 gt²);

 

solvin u gt tim of balls meetin as 2s.

height of balls meetin is =1/2 * 2 * g= 79.4m.

aftr this vel of the comb is half cuz mass has doubled(cons of momen) n this vel is upwards:

v_{1^{= 49 - 2g}}

resultant is 4.9 upwards. it takes 0.5 sec to com to rest.

height reached is now is 79.625m.

now this comb finally comes back to earth after 4.03 sec(usin frmula s=(1/2 g t²).

total time= 2+ 0.5+4.03=6.53s.