Mechanics Problem - tennis player hitting lob shot

[pelletpuss]

Mechanics Problem -- tennis player hitting lob shot

Homework Statement

A tennis player (1) located on the line at the back of the court attempts to “lob” his competitor. The tennis player (2) stands 2 meters behind the net facing player (1) who hits the ball in 0 (origin), 9 meters away from the net and at a height of 0.5m above the ground. The tennis ball travels with a speed Vo=12 m/s, with the velocity vector at an angle of 60degrees with the ground.
g = 9.81 m/s2

a. Determine the trajectory equation of the ball after being hit in plan(o,x,y)
(give all the details from where the equation comes from)

b. Give y = f(x) equation including the numerical parameters given.

c. Knowing that player 2 jumps, extending his tennis racket at the maximum, reaches a height of H=2.5m, do you think he will intercept the ball?

d. The rear tennis court line is 12m away from the net, will the ball be in? or will player1 fail?

Homework Equations

h=(V^2)/2g
s=ut+1/2gt^2
v=u+at
2h/g

The Attempt at a Solution

turned the vector into x & y components

x = 12cos60 = 6
y = 12sin60 = 10.39

t1 =
v=u+at
0 = 10.39-9.81t
10.39=9.81t
t=10.39/9,81
t = 1.06

found hmax

(10.39^2)/2*9.81
=5.50 + 0.5m = 6.00m

found t2

t2 = squareroot(2h/g)
t2 = 1.10

just stuck on wether that's correct or not, and how to go about b & c

 

[Simon Bridge]

Lessee... you appear to be neglecting air resistance.

How is the trajectory equation different from the y=f(x) equation?
Given y(x), you just want to find y(x=position of player 2).

You've chosen to find the equations of motion first - note: you really need to keep x and y as positions rather than speeds, as you have. So:

$\vec{r}(0)=(x,y,z)=(0,0.5,0)\text{ m}\\ \vec{v}(0)=(\cos\theta , \sin\theta, 0)u \text{ m/s}\\ \vec{a}(0< t<T)=(0,-g,0)\text{m/s}^2\\ \vec{r}(T)=(R,0,0)\text{ m}?\\ \vec{r}_{1}=(0,0,0)\text{ m}\\ \vec{r}_2=(11,0,0)\text{ m}$

Where T is the "total time of flight" ... does it matter if T is the time to hit the grund or the time to return to initial height (ie why not make the initial height at y=0m and the ground at y=-0.5m?)

Note $\sin(60^\circ)=1/\sqrt{3}$ ... don't round it off too soon.

You put t1=... what happens at time t1? You should say at the start - it helps you figure if your answers are correct later.
Same for t2=...

t1 appears to be the time to reach the maximum height. In "t=0.6" you lost the subscript and the units for some reason. You have to watch this.

t2 appears to be the time to fall to the ground again.

If all this is what you intended to calculate, then why are you stuck on whether this is correct or not?
What is it about the answers that bothers you?

I cannot tell if the approach is "correct" unless you tell me why you have calculated these values though.
i.e. $R=u(t_1+t_2)\cos\theta$ would be the distance from player 1 that the ball hits the ground but so what?

Personally, I'd make T the time it takes the ball to reach player 2 ... then $r(t)=(x(T),y(T),0)$.

You already know the equations for x(t) and y(t) right?