Cricket balls collision

The time for the ball to reach the top of its path Tt (where v=0) is given by v=u+aTt. So 0 = 30 -10Tt so Tt= 3.

After the second ball is launched the same amount of time most have passed for both balls to be at the point of collision and - more importantly - as they where launched with the same velocity this time most be half of the time it takes to reach the top (symmetrical accelerations) so the height of collision is given by s=ut - 1/2(at^2) = 30X1.5 - 1/2X10X(1.5^2) = 33.75 or 34 to the nearest meter.

The fact that the collision time occurs at 1/2 of the time needed to reach the apex can be seen as follows:-

Time to reach the top (Tt) is given from v=u+aTt as Tt=u/g (v=0 and a is -g)

The height it reaches (St) is then St =uTt+1/2aTt^2 = u(u/g) -1/2g(u/g)^2 = u^2/2g.

Now this height is equal to S1 + S2 the distances travelled by the balls when they collide at Tc, so (S1)+(S2) = (uTc - 1/2gTc^2) + (1/2gTc^2) = uTc and this distance is equal to the height that we just calculated as u^2/2g so uTc = u^2/2g or Tc = u/2g.

Now we can see that the ratio of the time to collision over the time to reach the apex is Tc/Tt = (u/2g)/(u/g) = 1/2. This applies to all values of g and initial velocities u where u is the same for both balls.

Plugging values into the above equations from our example (u=30m/s, g=10) we get.

h = 45, Tt = 3, Tc =1.5, S1 = 34m.

initial velocity ${ v }_{ 0 }=108\quad km/h\quad =30\quad m/s$

the maximum height of the ${ 1 }^{ st }$ ball is:

$h=\frac { { v }_{ 0 }^{ 2 } }{ 2g } =\frac { { \left( 30 \right) }^{ 2 } }{ \left( 20 \right) } =45\quad m$

consider $y$ is the height of the collision, so:

for the ${ 1 }^{ st }$ ball the height is $\left( 45-y \right)$

$\left( 45-y \right) =\frac { 1 }{ 2 } { gt }^{ 2 }$

$\left( 45-y \right) ={ 5t }^{ 2 }$

${ t }_{ 1 }=\sqrt { \frac { \left( 45-y \right) }{ 5 } }$

now, time for ${ 2 }^{ nd }$ ball to reach the collision position is equal to time on ${ 1 }^{ st }$ ball from the highest to the collision position. Now for the ${ 2 }^{ nd }$ ball:

$y={ v }_{ 0 }t-\frac { 1 }{ 2 } g{ t }^{ 2 }$

$y=30t-5{ t }^{ 2 }$

$y=\left( 30 \right) \left( \sqrt { \frac { \left( 45-y \right) }{ 5 } } \right) -5{ \left( \sqrt { \frac { \left( 45-y \right) }{ 5 } } \right) }^{ 2 }$

$y=\left( 30 \right) \left( \sqrt { \frac { \left( 45-y \right) }{ 5 } } \right) -5\left( \frac { \left( 45-y \right) }{ 5 } \right)$

$0=\left( 30 \right) \left( \sqrt { \frac { \left( 45-y \right) }{ 5 } } \right) -45$

$\sqrt { \frac { \left( 45-y \right) }{ 5 } } =\frac { 3 }{ 2 } \sqrt { \frac { \left( 45-y \right) }{ 5 } } =\frac { 3 }{ 2 }$

$\frac { \left( 45-y \right) }{ 5 } =\frac { 9 }{ 4 }$

$y=45-\frac { 45 }{ 4 } =33.75\quad \approx \quad \boxed{34\quad m}$