What are the steps to finding the angle and height for a tennis serve?

[physicshelpme]

Homework Statement

1. Consider tennis net which is 2 ft high at its center point; the service line is 20 ft from the net, and the baseline is 30 ft from the net. You have a bullet-like service that travels in a straight line, just clearing the net before hitting the service line.
a) What angle does the trajectory of this serve make with the horizontal?
b) At what height above the ground is the ball struck with the racket?

Homework Equations

1)i don't know how i would find an angle without given one? do i need to assume?
2)dont know where to start

The Attempt at a Solution

1)please help spent hours trying to solve this problem

 

[adjacent]

Welcome to PF :)
Can you draw a diagram at first?

 

[Dadface]

The ball cannot travel in a straight line. Ignoring air resistance it follows a parabolic path.

 

[physicshelpme]

heres what i drew and my thoughts

 

[ehild]

heres what i drew and my thoughts

It is not shown.

ehild

 

[physicshelpme]

picture

 

[physicshelpme]

so here what i think happened y=(vi x t)+1/2g x t^2
to find time
vi=0m/s y=2ft g=-9.8m/s so we can find t but we need to convert to meters

 

[physicshelpme]

i also did vf^2=vi^2 +2ay

vi=0m/s a=-9.8m/s y=2ft->.61meters

so vf=3.5m/s

 

[ehild]

The ball is hit with initial speed so great that its trajectory can be considered linear. You can ignore the g/2 t^2 term. As written in the problem text

You have a bullet-like service that travels in a straight line, just clearing the net before hitting the service line.

So the ball travels along a straight line between the upper point of the net and the service line, 20 ft away. What is the angle of the trajectory with the horizontal?

ehild

 

[physicshelpme]

how would i go about finding the trajectory with the horizontal?

 

[physicshelpme]

i think i got it now v^2=vo^2 +2-g(y-yo) vf=0m/s vi=? y=0 yo=2ft g=9.8?

 

[ehild]

Read the problem text again.

ehild

 

[physicshelpme]

i did, how do i find a tracetory angle?

 

[jackarms]

The problem says it travels in essentially a straight line from the start until the net, so the path lies directly along the horizontal. What then is the angle between the path and the horizontal?

 

[physicshelpme]

0? if its linear then it has to be 0?

 

[jackarms]

Yes, that's right. This problem seems rather odd though -- it's implying that it travels in a straight line and then suddenly starts to fall. If the problem is exactly as you typed it though, then an angle of 0 is the implied answer.

 

[physicshelpme]

thats exactly what he wrote i didn't change anything, b) so now we consider the height of the ball being struck
v^2=vo^2 +2-g(y-yo) y=0 landing of the ball in the y direction?

 

[jackarms]

Okay, but then you don't have either the initial or final velocities, and since the trajectory of the ball is atypical, your typical equations won't work unless you look at the path in intervals. Just looking at the straight line path from the start to the net, you have the final y as 2 feet, and y doesn't change over that path, so what is the starting y (the height from which the ball is hit)?

 

[physicshelpme]

yeah your right the equations are not going to work. but if we are only considering the object in the y direction. I am unsure what to do next, this guy never teaches the math in class just talks about definitions something we can simply read at home.
why is the service lime 20ft, and baseline 30ft mentioned in the problem is that to trick us? or used as sin(0), cos (0)

 

[rikblok]

The problem says it travels in essentially a straight line from the start until the net, so the path lies directly along the horizontal. What then is the angle between the path and the horizontal?

I don't follow. Just because it travels in a straight line doesn't mean the path has to be horizontal... just straight, right?

 

[CWatters]

This is the diagram I would draw. Only needs basic trig to work out the angle and H. You could also work out H using the law of similar triangles.

 

[physicshelpme]

yea that makes more sense c watters

 

[jackarms]

Ah yes, my bad. I glossed over the word "service" and just thought the ball was being hit somehow. I think that diagram is the best way to interpret the problem. So I guess it ended up being more geometry than anything...

 

[physicshelpme]

so for the tan-1(2/20) of the little triangle be the same as theta of the whole triangle?

 

[physicshelpme]

the angle is (.1) very small

 

[physicshelpme]

i got 5.1 as the height

 

[CWatters]

Check the angle. I got 5.7 degrees and exactly 5 ft for the height.

Tan^-1(2/20)=5.7

H=(2/20) * 50 = 5 using the similar triangles rule

 

[CWatters]

Just for completeness this is how the similar triangles rule works...

The angle is the same for the small and large triangle so:

Tan(θ) = 2/20
Tan(θ) = H/50

2/20 = H/50

H = (2/20) * 50 = 5ft