INSTRUCTOR: An archer fires an arrow at an angle of 40 degrees above the horizontal with an initial speed of 98 meters per second. The height of the archer is 171.5 centimeters. Find the horizontal distance the arrow travels before it hits the ground.

Now, let's notice some information we have. First, theta is equal to 40 degrees above the horizontal. The initial speed, v0, is equal to 98 meters per second. And we also know the initial height. The archer is 1.715 meters tall. That will be useful in a moment.

Now, we know there's an equation that can tie these things together. The position function s of t equals v0 t times cosine theta i plus parentheses v0 t multiplied by sine theta minus 1/2 gt squared close parentheses, j, will give us the position that we want. Now, filling our information in 40 degrees as theta and initial velocity of 98, this means that our position function, s of t, equals 98t cosine of 40 degrees i plus parentheses 98t sine of 40 degrees minus 1/2 times 9.8t squared close parentheses j. That is the position of the arrow.

Now, what we need to find is when the vertical component, that would be our sine, the j component, we want to find when that component is 1.715 meters below the initial height. So let's make this into a quadratic. We've got that it would be a negative 4.9, think about the vertical component, negative 4.9 squared plus 98 sine of 40 degrees t. When is that equal to negative 1.715?

Now, I'm going to add 1.715 to both sides and use quadratic formula to show that t is equal to negative 98 sine of 40 degrees plus or minus the square root of parentheses 98 sine 40 degrees close parentheses squared, minus multiplied by negative 4.9 multiplied by positive 1.715. Square root of that, and all of that is over 2 multiplied by negative 4.9.

Now, there will be two solutions to this approximate solutions. The one we care about is the positive value. So t is approximately 12.8829 seconds. That is when it hit the ground. We need to evaluate that into our function here.

That would be s of 12.8829. Let me just go in and write that, 12.8829 is approximately equal to 967.15 comma 0. And so the horizontal distance it travels is approximately 967.15 meters.