What is the position vector of the ball in a tennis serve?

[Kawakaze]

Homework Statement

A player stands with his feet at the point O and serves the ball from a point Y at
a height of 2.4000m vertically above O. Assume that the ball then travels in a
straight line directly over the net at the midpoint A, which is at a height of
0.9144 m, before bouncing at the point X.

Determine the position vector of any point on the line Y A, and hence find
the position vector of the point X. Deduce that the ball cannot land in the
shaded area of the court.

Homework Equations

Excuse my lack of vector notation, i,j and k should be in bold :)

$$t=(1-s)p+sq$$

The Attempt at a Solution

$$t=(1-s)(0i, 0j, 2.4k) + s(11.8872i, 4.1148j, 0.9144k)$$

$$t=11.8872si + 4.1148sj + (2.4 - 1.4856s)k$$

I know I have to rearrange somehow and find i and j when k = 0, but that's about it.

 

[Delphi51]

In t(1-s)p + sq what does each variable mean?

It seems to me it is useful to draw a side-view sketch showing the height of the ball (y) as it travels across the court. I think the slope of this line must be found to predict the distance the ball travels before hitting the floor.

 

[Kawakaze]

Hi Delphi,

In the formula, s is in between 0 and 1, p is the starting point, and q is the finishing point. this i am told is the equation for a straight line vector, s=0 is the start and s=1 is the end. I assume be rearranging somehow, i can get i & j at any wanted value for k.

 

[Delphi51]

Okay, makes sense now. Your vector
$$11.8872si + 4.1148sj + (2.4 - 1.4856s)k$$
represents the path of the ball when different values of s are used.
You need to find the value of s that makes the k component zero:
2.4 - 1.4856s = 0
for the point when it has height zero.
Then you can use that value of s in the i and j components to find the position on the floor where it hits.

 

[Kawakaze]

Hi Delphi,

Thanks for the help, I was trying that approach but forgot to take the K out :)

Solved, thanks again!