Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 4040^\circ above the horizontal.
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Final Answer

18 m/s18\textrm{ m/s}

Solution video

OpenStax College Physics, Chapter 3, Problem 42 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. A soccer player kicks a ball which is on the ground at the level of zero meters initially. With an angle of 40 degrees and the goal is 30 meters away. And the ball just barely passes over the goal posts at 2.4 meters in the air. So the trajectory is kind of like this. Well, you get the idea anyway. It goes along some parabolic path such that here is the height of 2.4 meters when it's 30 meters away from where it started. Okay, and we have to figure out what is the initial speed of the ball. So, we’ll begin by writing down some of the kinematic equations that we know. And for the x-direction, we'll look at equation 53 from chapter two. I will say that the final x-position equals the initial x-position, plus the initial x-component of its velocity, multiplied by time. Plus one half times its x-acceleration, times time squared. Now we don't know what the time is, but let's just put that thought to the side for a moment. Let's begin by just saying that the initial x-position we’ll define as zero. And there is no acceleration in the x-direction. And so both of these terms disappear. And so we have equation number one; which is that its final x-position is the initial x-component of the initial velocity, multiplied by time, whatever that time is. So that's the best we can do at this point for the x-direction. We have two unknowns, and that's a bit tricky. But let's build our solution here by considering the y-direction. We’ll use the same equation and we’ll say that the initial y-position is zero, and the acceleration in the y-direction is negative g. Where we've you know defined a coordinate system in the usual way with up to be positive and right positive. So we make those substitutions. This term is zero and this term becomes minus g in place of plus a y. And then this time I mentioned is an issue, and we can get rid of it by rearranging equation one. And making version b out of it; equation one b by solving for time. And then substituting for time into equation two, and that will make the time factor disappear. So I have divided both sides by v naught x here to solve for t here. And then that's getting substituted in place of t in this equation number two. Which we’ll call equation two version b. So the y-position then is the y-component of the initial velocity, multiplied by x over the x-component of the initial velocity. Because that's what I'm substituting in place of t, minus one half g x over v naught x squared. Now. that is not the whole story, because this equation has two things that we don't know; It has the y-component of the initial velocity and the x-component of the initial velocity that we don't know. We do know y, that's 2.4 meters, the height of the goalpost. We do know x, that's 30 meters and we know g. So, this means that we need to have a yet another equation to deal with the fact that there are two unknowns. And before we get to the story here, we're going to use tangent theta to solve this issue. But one principle to keep in mind when you have this system of equations. And by that I mean you have several different equations that all describe the same physical situation. That's called a system of equations. When we have a system of equations, you need to have as many equations as you have unknowns. And so when I wrote down equation one, we can see that there are two unknowns. And when I wrote down equation two, we can see that we've introduced a third. This t is already accounted for here. So there are three unknowns v naught x, t, and v naught y. And when you have three unknowns, it means you have to have three equations to solve for any of the factors. So to solve the system of equations, you need to have as many equations as you have unknowns. So and here we have equation number three. We have three and unknowns and so we need three equations. So we can say that tangent of theta, the launch angle here is, well, I’ll draw the velocity triangle here to make it more clear. Here’s the x-component v naught x. And here's the y-component v naught y, and here’s the angle theta. So we can see that the tangent of this angle is the opposite, v naught y, divided by the adjacent v naught x. And so that's what I'm writing here and that's useful because we have v naught y divided by v naught x here. And so this can all get replaced by tangent theta. And so that's what I've done in equation number two version c. Which is replacing v naught y over v naught x with tan theta. So y then is tan theta times x, minus one half g x squared over v naught x squared. Where you know applied the exponent to both of these factors. Well this is great because now we have an equation that has only one unknown. We know everything else in here, we’re given theta, we’re given y. We’re given x, and we know g. So we just solve the equation for this. So, I move this term to the left hand side which makes it positive; So we have g x squared over two v naught x squared. Then I move this y to the right hand side which makes it minus y. o we have g x squared over two v naught x squared equals x tan theta, minus y. Then, I raise both sides to the exponent negative one just to get this unknown into the numerator. Because it's kind of inconvenient having it in the denominator. And this also is raised to the power of negative one. And keep in mind that this whole thing can be written as over one. So exponent negative one means flip the fraction. And so we have two v naught x squared over g x squared. Equals one over x tan theta, minus y. Then multiply both sides by g x squared over two. We end up with this line here, v naught x square, equals g x squared over two times x tan theta minus y. And then take the square root of both sides. And this x squared, can be taken out of the square root sign and written as x here. And so we have v naught x equals, strictly speaking plus or minus, but we know that the ball is moving to the right and that's our positive direction. So it's plus. So we have plus x, square root of g over two x tan theta minus y. So that is 30 meters, times square root of 9.80 eters per second squared, over two times 30 meters. Times tan of 40 degrees minus 2.4 meters, giving an x-component of its initial velocity of 13.9158 meters per second. Then we have to find the y-component of its initial velocity and we’ll use the tangent equation to do that. So tan theta is v naught y over v naught x, as we mentioned before. And so we’ll solve this for v naught y, by multiplying both sides by v naught x. So we have v naught y is v naught x, times tan theta. So that's 13.9158 meters per second times tan 40. Which is 11.6768 meters per second. Now the initial speed then is the Pythagorean sum of these x and y-components. So that’s a square root of v naught x squared plus v naught y squared. Which is a square root of 13.9158 meters per second squared plus 11.6768 meters per second squared. Which gives an initial speed of 18 meters per second.


This is how I approached it, and I got a different answer: solve first for V0V_0 of y using V2V^2 = (V0)2(V_0)^2+2a(Δ\Deltay). I assumed VyV_y =0 since the problem says the ball 'just passed over the goal', so it reached the vertical max where velocity = 0. Plugging in 0= (V0)2(V_0)^2+2(-9.8)(2.4) and solving for (V0)2(V_0)^2, I got 6.86 m/s as the initial velocity of y. I then plugged that initial velocity into the formula v= (V0)(V_0)+at to get time, which I know is constant for x and y. That gave me 0.7 seconds. I then used time to plug it into the following formula to solve for the (V0)(V_0) of x: (Δ\DeltaX)=(V0)(V_0)t+1/2(at). Since I know acceleration in the horizontal is 0, that meant that formula became (Δ\DeltaX)=(V0)(V_0)t. Since I'm given (Δ\DeltaX) (30 m) and had previously found t of 0.7 seconds, that gave me (V0)(V_0) = 42.9 m/s... and then using pythagorean theorem, i got c2c^2 of 43.4 m/s. What am I doing wrong here?

Hi becky, thank you for the question. The issue is the assumption that Vy=0V_y = 0 when the ball is above the goal. 'just passed over the goal' could occur while still having a vertical component of velocity. Making this assumption presumes that the goal is at the half way point of the ball's range.
All the best,


Why wouldn't we use the -9.8m/s^2 value for g?

Also, for some reason when I plug those values into my calculator my answer if off by ~.15.

Hi kkm, thank you for the question. The assumption is that gg represents the magnitude of the acceleration due to gravity. gg is always positive in other words. When using it in the equation to substitute for the acceleration due to gravity, we introduce a negative sign in front of it to indicate it's direction downward at 1:53 in the video (since we've made the conventional choice of down being the negative direction).
Hope this helps,