## Hitting a moving target below when launched 100 m above at a 45 degree angle

i'm having difficulty trying to figure out a problem i thought of.

a projectile is launched upwards at a 45 degree angle 100 m above a surface. below is a target whose horizontal distance is 5 m from the launch point. just as the projectile is launched, the target moves horizontally at a constant 9 m/s.

what would the velocity of the launch have to be in order to hit the target below?

so far i have come up with:

5 + 9t = vcos45t

for t = time i have:

(see attachment)

but i'm having difficulty knowing how to incorporate it properly into the equations.

i've reduced the time to:

√([.02548v2 + 100]/4.9) + (.07215v)

is there an easier way to go about solving this?

any help would be greatly appreciated.

thank you
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 Mentor I don't think there is an easier way. I would try to use g instead of its numerical value and the analytic value for sin(45°) as long as possible, that avoids decimal numbers.

 Quote by mfb I don't think there is an easier way. I would try to use g instead of its numerical value and the analytic value for sin(45°) as long as possible, that avoids decimal numbers.
groovy!

i came up with a velocity of 13.985 m/s at a time of 5.638 seconds

thanks for your help...now i can sleep soundly :-)