Question
Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle $\theta$ below the horizontal. The baseline from which the ball is served is 11.9 m from the net, which is 0.91 m high. What is the angle $\theta$ such that the ball just crosses the net? Will the ball land in the service box, whose service line that is 6.40 m from the net?

$\theta = 6.1^\circ$

Yes, the ball will land within the service box.

Solution Video

# OpenStax College Physics Solution, Chapter 3, Problem 37 (Problems & Exercises) (15:37)

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Submitted by elnazassadpour on Sat, 03/16/2019 - 06:42

You are incredibly good at explaining the solutions. Thanks so much! Just a quick question. How come we can't find theta by just doing d = squareroot of dy^2 + dx^2. and then using the equation theta = tan-1 (dy/dx), basically just using distances that we know to find the angle, rather than velocity? There are other questions, where it's just distance that's used.

Submitted by ShaunDychko on Tue, 04/02/2019 - 09:52

Hello, sorry for taking so long to get back to your question. It was spring break here. Thanks for the nice feedback and I'm really glad the solutions are helping with your studies.
Let's keep in mind that theta is an angle to do with velocity. It's the angle between the 'x' and 'y' components of the velocity. It's possible to use distances as a proxy for velocity only when both distance components are proportional to their respective velocity components, in which case the triangles made from resultants and components will be similar. I use the word similar in the technical math sense in that the triangles will have the same angles since they have equal ratios of corresponding sides. However, in this particular question, the y-component of displacement will not be proportional to the y-component of initial velocity since there is vertical acceleration due to gravity. Displacement can be used to find theta because the acceleration interferes with that strategy.

Submitted by elnazassadpour on Sat, 03/16/2019 - 07:10

Do you suggest any resources to become more familiar with trigonometric identities, such as the ones used in this question.

Submitted by elnazassadpour on Sat, 03/16/2019 - 09:01

Why can't we use range =V2Sin2theta/a to solve for the second portion of the question asking for how far the ball will travel? Also for the second part of the question, why can't we use directly solve for x by manipulating the first equation so that t=x/Vcostheta, rather than solving for time and them substituting it into the equation?

Submitted by ShaunDychko on Tue, 04/02/2019 - 09:55

The range formula was derived with an assumption that is important to keep in mind: the initial height and final height must be the same. The tennis ball starts high, and ends low, so the range formula doesn't apply.

Submitted by kelseyrandlett on Sat, 03/28/2020 - 12:19

Hi there! In the first step of the question, you gave the original equation for yn=. In that first equation, you wrote +Vot, but when you substituted with Sin theta, you changed it to a subtraction. I was just wondering why this was done. Thank you!

Submitted by ShaunDychko on Mon, 03/30/2020 - 13:34

Hi kelseyrandlett,
Thanks for the question. The first equation is written generally, with no specific problem in mind. The substitution with Sin theta is done specifically for this particular problem, and from the figure we can see that the vertical component of the initial velocity is downwards, which is the negative direction. Putting a negative sign in the equation with this substitution is a personal preference approach to dealing with the downward direction of the velocity. The negative could alternatively be dealt with at the stage where you replace variables with numbers, and we could have substituted a negative number in for the velocity and arrive at the same result. If find that confusing though. My preference is to substitute magnitudes for velocities, and make direction indicated by signs in the equation.
Hope that helps,
Shaun