Vertical projectile motion

The first ball is in the air for $1.2$ seconds (standard projectile calculations - lots of different ways to get this). During this time, $5$ other balls are thrown; the first ball must meet each of these on its way down.

Here they are:

First, we determine, for how long is the first ball in the air.

For this, we can use SUVAT:

$u = 6 \frac {m}{s}$

$a = g = -10 \frac {m}{s^2}$

t = ?

s = 0 (the displacement is 0 when it arrives back to the same height it started)

After this, we can choose the following SUVAT equation:

$s = ut + \frac {1}{2} a t^2$

$0 = 6t - 5t^2$

$0 = t(6 - 5t)$

This gives us t = 0 second (the time when we throw the ball upwards) and t = 1.2 seconds (when it is at the same level it started. During this time, it meets 5 balls thrown (at t=0.2, 0.4, 0.6, 0.8 and 1.0 seconds) subsequently, still above its initial position.

Hence, our answer should be: $\boxed {5}$

The vertical position of the first projectile may be given by:

$\displaystyle y(t) = v_o t - \frac{1}{2}g t^2$

The vertical position of the nth projectile can be given by:

$\displaystyle Y(t,n) = v_o \left( t - \frac{2}{10}n \right) -\frac{1}{2}g \left( t - \frac{2}{10}n \right) ^2$

The projectiles must meet in the during the time the first ball is in flight. Solving for the time duration of the first projectiles flight:

$\displaystyle y(t) = 0 \rightarrow t= \left \{0, 2 \frac{v_o}{g} \right \}$

Now equate the positions, and eliminate the variable $t$ (time) from the solution above:

$\displaystyle y(t) = Y(t,n)$

$\displaystyle \frac{2}{10}v_o n - \frac{2}{100}g n^2 = 0$

$\displaystyle n = \left\{0,6\right\}$

The second projectile being fired at n=1 we remove 0 as a valid solution, and the last solution occurs at n=6 which coincides with the duration of the first projectiles flight time, in other words position zero. Removing that final solution means the first projectile meets with 5 other projectiles at some position greater than zero.

The Answer: 5