Question 295516
The flare is acting under the force of gravity with an initial upward velocity.
{{{ a=dv/dt=-g }}}
{{{ v=dy/dt=-gt+C1}}}
{{{ y=-(1/2)gt^2+C1*t+C2}}}
where g is the gravitational constant (9.8 m/s^2), y is the flare's position from the ground (y=0 on the ground and moves positively upwards), t is time. C1 and C2 are constants that you solve for using your initial conditions.
{{{dy/dt}}} is velocity, the first derivative of position. 
{{{dv/dt}}} is acceleration, the second derivative of position, first derivative of velocity.
You know the initial velocity (t=0).
{{{ -9.8t+C1=294}}}
{{{ -9.8(0)+C1=294}}}
{{{ highlight( C1=294)}}}
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You also know the initial position (t=0). The flare is on the ground (y=0).
{{{  y=-(1/2)gt^2+C1*t+C2}}}
{{{ -(1/2)(9.8)(0)^2+294(0)+C2=0}}}
{{{ highlight( C2=0)}}}
Now you have the complete equation for the flare's position as a function of time.
{{{ highlight_green(y(t)=-4.9t^2+294t)}}}
Solve for t when the flare is at the balloon position,{{{y=2450}}}.
{{{ 2450=-4.9t^2+294t}}}
{{{ 4.9t^2-294t+2450=0}}}
Use the quadratic formula,
{{{ t=(294 +- sqrt(294^2-4(2450)(4.9)))/(2(4.9))}}}
{{{ t=(294 +- sqrt(86436-48020))/(9.8)}}}
{{{ t=(294 +- sqrt(38416))/(9.8)}}}
{{{ t=(294 +- 196)/(9.8)}}}
{{{ t1=(490)/(9.8)=10}}}
{{{ t2=(98)/(9.8)=50}}}
At time {{{t=10}}} seconds, the flare passes the balloon going up and at {{{t=50}}} seconds the flare passes the balloon going down. So {{{40}}} seconds after the flare passes the balloon going up, it will pass the balloon going down.