F the arrow is released from a height of 1.00 m above the ground and hits the apple 0.520 s later,

William Tell is shooting at an apple that hangs on a tree. The apple is a horizontal distance of 20.0 meters away and at a height of 4.00 meters above the ground.

If the arrow is released from a height of 1.00 m above the ground and hits the apple 0.520 s later, what is the arrow’s initial velocity?

Please help me with solving this.

Horizontal speed = 20/0.52 = 38.46 m/s

Using s = ut + ½ at^2 ,

( 4 -1) = v*0.52 – 4.9*0.52^2 where v is the initial vertical speed.

v = 8.32 m/s .

Initial speed = √ (38.46^2 + 8.32^2) = 39.35 m/s

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So the arrow isn't shot completely horizontally because we already know the apple is 3 feet above the arrow. So we know the arrow is shot at an angle, which means it has some velocity in the x direction and some in the y direction. And we can find out the x by simple kinematic equations, we take the equation:

D=vi*t+(1/2)at^2 , so to find out your vi in the x direction you plug in what you know about the x direction as shown below:

Dx=vix*t+(1/2)axt^2 , and we know that acceleration in the x direction is always zero , so you have
Dx=vix*t , and we know the time and the distance in the x direction so then we have vi in the x direction

Then we do the same for y direction ,

Dy=viy*t+(1/2)ayt^2 and again we know everything except viy which we can solve for

Once we have vi in both directions we can do some Pythagorean theorem to find the resultant velocity. And then your finished

Here are some things that could help you.
Dy=3
Dx=20
T=same in both directions
Ay of course is gravity which is -9.8 the negative is important