# Homework Help: Finding the general highest spot-balistick

1. Nov 23, 2012

### Lenjaku

1. The problem statement, all variables and given/known data
a ball is thrown in angle α from height H , moving horizontal distance D.
express the highest spot it reaches using the parameter D.

2. Relevant equations
r=r0+v0(t)t+0.5gt2

3. The attempt at a solution
My question is why I can;t simply find the highest spot using the vectors without using the equation?
The velocity vector is integrated from the acceleration so it already contains the acceleration so I dun get why I get a wrong answer.
Here is what I did:

a=(0,-g)

v=∫a +v0=>
v=(vcos(α),vsin(α)-gt)

now I found t by saying highest spot is when vy=0
so I get

vsin(α)-gt=0 =>
t=( vsin(α) )/g

Now here is what I am supposedly doing wrong:

The position vector in any given time is:
∫v+r0 =>
r(t)=(-vsint(α)t,H+vcos(α)t-gt2)

Ok now why can't I simpy
calculate it as r(t=( vsin(α) )/g) ?
Why I have to use the equation r=r0+v0(t)t+0.5gt2?

I am looking for the D it will reach in the given time (which I calculated as the time it takes the ball to reach it's highest point).
The acceleration is already in account so what am I missing?

the next steps I'd take are saying:
rx=D and finding t using D.
ry=0. using the t I just found (all this to get D involved). (here again I would just place t in teh vector but the answer actually contains the equation of r=r0+v0(t)t+0.5gt2 ) why can;t I use it without it?

And eventually using the new ts I just found (which contain the D I need)

Place it again in the vector which expresses highest height.

I know in a specific t it is in a specific position , I wanna find the t it reaches with specific speed....why do I need the equation if I have specific points?
There is no problem using for instance the place it reaches in t=3.
I dun get it.

Last edited: Nov 23, 2012
2. Nov 23, 2012

### ehild

Your last formula is wrong. Otherwise, you can substitute tmax for t in r(t), to find the maximum height.

ehild.

Last edited: Nov 23, 2012
3. Nov 23, 2012

### Lenjaku

I think I figured out...
Using the equation or using the vector itself should be the same.
the equation was adapted from the vector's integration the missing 0.5 that confused me was missing because I forgot that when integrating pow u also add it as friction.
(sorry I am having hard time translating math terms,I meant 2t→ (2t2)/2
and not simply 2t2...the rest of the logic may be off due to my lack of concentration I keep loosing myself and mixing things/forgetting things etc -.-

Anyway I think the problem is solved I just have to start over.

4. Nov 23, 2012

### ehild

There is the time when the ball reaches the highest point. It is t1=(vosinα)/g.
There is the "time of flight" t2, from start to the instant when the ball reaches the ground: y(t2)=0. You get the maximum horizontal distance with
D=t2 vo cosα.

Find ymax in term of D (and alpha and H).

ehild