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## Homework Statement

An archer launches an arrow from coordinates ##(0, 0)## at an angle ##\alpha## and with an initial velocity ##v_0##. There's a target located ahead of the archer and the center of that target is at coordinates ##(d, h)##. At what ##v_0## and at what angle ##\alpha## does the archer need to launch his arrow to hit the center of the target

**perpendicular to its surface**?

## Homework Equations

Kinematic equations

## The Attempt at a Solution

We have that:

##v_x = v_0\cos(\alpha)##

##v_y = v_0\sin(\alpha) - gt##

For the arrow to hit the center of the target perpendicular to its surface, it must have a vertical velocity of zero when it reaches the target. So the following conditions must be met:

1) By the time the arrow travels a distance ##d## in the ##x## direction, its vertical velocity component should be equal to zero.

2) By that same time above, the arrow should be located at height = ##h##.

So we have:

##d = v_0\cos(\alpha)t_i \rightarrow t_i = \frac{d}{v_0\cos(\alpha)}## ..................................... (1)

##0 = v_0\sin(\alpha) - gt_i \rightarrow t_i = \frac{v_0\sin(\alpha)}{g}## ............................... (2)

##\frac{d}{v_0\cos(\alpha)} = \frac{v_0\sin(\alpha)}{g} \rightarrow v_0^2\sin(\alpha)\cos(\alpha) = dg## ................. (3)

Where ##t_i## is the time of impact.

I'm at a loss as to how to proceed from here. I'm assuming I need to use the displacement equation in the vertical direction: ##h = v_0\sin(\alpha)t - \frac{g}{2}t^2##, but I am not sure what to put into this equation. Do I put in the ##t_i## from equation (1) or from equation (2)? Can I use

*both*values of ##t_i## simultaneously (insert one under ##t## and another under ##t^2##) since they are supposed to be the same thing? I tried doing this and I get an algebraic expression that I'm unable to process. For example, here's what happens when I use ##t_i## from equation (1):

$$h=\frac{v_0\sin(\alpha)d}{v_0\cos(\alpha)} - \frac{gd^2}{2v_0^2\cos^2(\alpha)} \rightarrow h = \frac{sin(\alpha)}{\cos(\alpha)} - \frac{gd^2}{2v_0^2cos^2(\alpha)}$$

But this equation doesn't account for the fact that at time ##t_i##, the vertical velocity needs to be zero. How do I proceed?