 Here we have a typical projectile motion problem. You have a projectile that is shot at a certain angle with some initial velocity, and then due to gravity follows a parabola and lands on a target sometime later. In this problem here, we're given that the initial velocity here had a magnitude of 40 meters per second, and we're given that the target lands four seconds later at exactly the same height from which it was shot from. What we're looking for is what is the angle, and what is the distance in X that the projectile traveled. Now, first of all, what we can do, we can split up that initial velocity in its X and Y component. So we have VX initial is V initial times the X side is coside of the angle. Here we have VY initial, V initial times sine of the angle. Also we have the X position as a function of time is the initial position, so in this case zero, plus VX initial times time, plus one-half exploration t-square, there is no exploration in X direction, the exploration is only Y direction, so that's it for this one here. So for S, Y as a function of time is zero, I'm starting at a height zero, plus VY initial times time, plus one-half exploration time script. Now in my problem here, I'm looking for the angle, and I'm looking for the delta X. The delta X, if I don't my X component here, I can easily find from this equation over here. The problem is for this equation to be solved, I need to know the alpha, that means for the moment I cannot solve from the X, so what I'm going to be doing, I'm solving from the Y equation, hoping that gives me the alpha and then plug it into the X equation. Let's start with the Y equation. The Y equation, I know that when it lands again, I have zero, must be equal to V initial times time, so V initial times time, so V initial times sine, alpha times time, plus one-half exploration time script. So in my case, V initial was 40 meters per second times sine of the unknown angle times four seconds, then for the exploration I'm going to be rounding minus tens, I'm going to use minus five meters per second square times four squared is sixteen, four squared seconds squared. Now all I have to do is solve this for alpha. So I have five times sixteen squared, that I take on the left, and I divide it by the forty times four, and then isolate for alpha. So if I look at the units, I have meters per second squared times second squared gives me meters, and down here I have meters per second and seconds also gives me meters, that cancels out, so I get zero unit. Sine of alpha is zero point five, therefore alpha is our all-time favorite angle of thirty degrees. So I figured out my first answer, which is thirty degrees. Now that I know my angle, I can solve for my exposition, so I'm just going to make some space. Okay, now that we know that alpha is thirty degrees, part B is quite simple, we simply plug it into the equation here. So we have S as a function of time at four seconds, will be my V nation X, so forty meters per second times cosine of thirty degrees times the time, which is the four seconds, meters, so if you round to one thick, or two thicks, 140 meters.