Question

A ball is thrown vertically upwards with a velocity of 29.4 m/s. After 3 s, another ball is thrown upwards from the same point with a velocity of 19.6 m/s When and at what height will the two balls collide?

Answer 1

$\large\underline{\underline{\sf Given:}}$

• Velocity of first ball $\sf{(u_1)}$ = 29.4 m/s

After 3 sec Another ball is thrown upward

• Velocity of second ball $\sf{(u_2)}$ = 19.6m/s

$\large\underline{\underline{\sf To\:Find:}}$

• Time of Collision of two ball $\sf{(t_1)}$ = ?

• Height of Collision = ?

$\large\underline{\underline{\sf Solution:}}$

Assume ,

direction towards downward direction = -ve

direction towards upward direction = +ve

g = - 9.8 m/s²

According to third equation of motion

$\large{\boxed{\sf s=ut+\frac{1}{2}at^2}}$

Now ,

Position of first ball :-

$\large{\sf s_1=29.4t_1-4.9t_1^2}$

Position of second ball :-

$\implies{\sf s_2=19.6(t_1-3)^2-4.9(t_1-3)^2}$

As the balls colloid at

$\implies{\sf t=t_1\:s\:and\:s_1=s_2}$

•°• $\implies{\sf 29.4t_1-4.9t_1^2=19.6(t_1-3)-4.9(t_1-3)^2}$

$\implies{\sf 6t_1-t_1^2=4t_1-12-t_1^2+6t_1-9}$

$\implies{\sf 4t_1-21=0}$

$\implies{\sf t_1=\dfrac{21}{4} }$

$\implies{\sf t_1=5.24\:s }$

$\large\red{\boxed{\sf t_1=5.24\:s}}$

On putting value of $\sf{t_1}$ in $\sf{s_1}$ :-

$\implies{\sf s_1=29.4×5.24=4.9(5.24)^2 }$

$\implies{\sf s_1= 154-4.9×27.45}$

$\implies{\sf s_1=154-134.54 }$

$\implies{\sf s_1=19.4m}$

$\large\red{\boxed{\sf s_1=19.4\:m}}$

$\Large\underline{\underline{\sf Answer:}}$

Hence ,

$\large\underline{\underline{\sf Time\:of\: Collision\:of\:two\:ball\:is\:5.24\:s}}$

$\large\underline{\underline{\sf Height\:at\:which\:both\:ball\:}}$

$\large\underline{\underline{\sf Colloid\:is\:19.4\:m}}$

Answer 2

Answer:

Velocity of first ball \sf{(u_1)}(u

1

) = 29.4 m/s

After 3 sec Another ball is thrown upward

Velocity of second ball \sf{(u_2)}(u

2

) = 19.6m/s

\large\underline{\underline{\sf To\:Find:}}

ToFind:

Time of Collision of two ball \sf{(t_1)}(t

1

) = ?

Height of Collision = ?

\large\underline{\underline{\sf Solution:}}

Solution:

Assume ,

direction towards downward direction = -ve

direction towards upward direction = +ve

g = - 9.8 m/s²

According to third equation of motion

\large{\boxed{\sf s=ut+\frac{1}{2}at^2}}

s=ut+

2

1

at

2

Now ,

Position of first ball :-

\large{\sf s_1=29.4t_1-4.9t_1^2}s

1

=29.4t

1

−4.9t

1

2

Position of second ball :-

\implies{\sf s_2=19.6(t_1-3)^2-4.9(t_1-3)^2}⟹s

2

=19.6(t

1

−3)

2

−4.9(t

1

−3)

2

As the balls colloid at

\implies{\sf t=t_1\:s\:and\:s_1=s_2}⟹t=t

1

sands

1

=s

2

•°• \implies{\sf 29.4t_1-4.9t_1^2=19.6(t_1-3)-4.9(t_1-3)^2}⟹29.4t

1

−4.9t

1

2

=19.6(t

1

−3)−4.9(t

1

−3)

2

\implies{\sf 6t_1-t_1^2=4t_1-12-t_1^2+6t_1-9}⟹6t

1

−t

1

2

=4t

1

−12−t

1

2

+6t

1

−9

\implies{\sf 4t_1-21=0}⟹4t

1

−21=0

\implies{\sf t_1=\dfrac{21}{4} }⟹t

1

=

4

21

\implies{\sf t_1=5.24\:s }⟹t

1

=5.24s

\large\red{\boxed{\sf t_1=5.24\:s}}

t

1

=5.24s

On putting value of \sf{t_1}t

1

in \sf{s_1}s

1

:-

\implies{\sf s_1=29.4×5.24=4.9(5.24)^2 }⟹s

1

=29.4×5.24=4.9(5.24)

2

\implies{\sf s_1= 154-4.9×27.45}⟹s

1

=154−4.9×27.45

\implies{\sf s_1=154-134.54 }⟹s

1

=154−134.54

\implies{\sf s_1=19.4m}⟹s

1

=19.4m

\large\red{\boxed{\sf s_1=19.4\:m}}

s

1

=19.4m

\Large\underline{\underline{\sf Answer:}}

Answer:

Hence ,

\large\underline{\underline{\sf Time\:of\: Collision\:of\:two\:ball\:is\:5.24\:s}}

TimeofCollisionoftwoballis5.24s

\large\underline{\underline{\sf Height\:at\:which\:both\:ball\:}}

Heightatwhichbothball

\large\underline{\underline{\sf Colloid\:is\:19.4\:m}}

Colloidis19.4m