What Was the Initial Speed of the Horizontally Shot Arrow?

[1234567890]

Help finding the initial speed please !

Homework Statement

An archer shoots an arrow horizontally at a target 12 m away. The arrow is aimed directly at the center of the target, but it hits 52 cm lower. What was the initial speed of the arrow? (Neglect air resistance.)

Homework Equations

well I am just dumb founded on this problem could some one please help me.
the equation i was given is : y= h - .5at^2 and x=vo*t so we then know t= x/vo

second part of the equation is y=h -.5 a(x/vo)^2
then vox* the square root of (a/2)*(x^2/h-y)

The Attempt at a Solution

the problem I am having is finding all the right data to complete the problem. We know x = distance which then = 12 and h(height) = 52cm or .52m

so in the first equation when it says y=h-.5at^2 and x=vo*t we need to find t so we can subsitute it for t^2 in the 1st equation. so the equation is t=x/vo and we get t=12/vo i don't need the answer to the problem i just need help finding what t is so i can solve it for myself. can you please explain how to find vo or whatever i need to find to start this problem. I've been working on it for a while now and i can't seem to come up with any answers.

Thanks a lot to anyone who can be a hand

 

[Sakha]

You know that the arrow accelerate downwards 9.8m/s² (g), so using your distance formula, you could get the time it takes to the arrow to travel 52cm downwards.
That same time is the time that the arrow, with speed v travels 12m in direction to the target.

 

[1234567890]

thanks man i got it