Projectile launch at elevated height on angle

rahrahrah1

1. The problem statement, all variables and given/known data

a baseball player hits a home run. The ball's velocity as it leaves the bat is 45m/s at an angle of 36.9° above the horizontal. At the point of impact with the bat, the ball is 1 m above ground level, and as it clears the fence it is back down to 11 m above ground level. How far from home plate is the fence at this point?

2. Relevant equations
quadratic formula
linear motion equations
basic sin and cos
v=d/t

3. The attempt at a solution

so I started with finding the Vv and the Vh velocties
vh= 45cos6.9= 36.0
Vv= 45sin36.9 = 27.0

then I organized the information I had into two seperate headings
Vv
v1: 27 m/s
v2:*27.4m/s
a: -9.8m/s
t:*5.5
d: -1m

and Vh
vave: 36m/s
t:*5.5
d:*198m

*I found the time using the quadratic formula and came up with 5.5 seconds
* I calculated the total distance the ball travelled using the formula d = v×t and got 198 m
* calulated the v2 for the Vv to be 27.4 m/s using linear motion equation
I'm thinking i need a colum for the fence distance now, however, this part is slightly iffy.

Vh for fence
vave:
t:T
d: 198-D

Vv for fence
v1:
v2:
a:-9.8m/s
t:T
d: -11m

At this point I've hit a dead end.

emailanmol

The question states the ball is at 11m above the ground

So displacement is 10m.
Calculate your time again.(its not right)

After you have time
You will need to multiply it by v(h)*t to get d.(as you already did)

Thats all.thats your answer

I dont understand what (and why)you are doing in the latter part of your post.

rahrahrah1

Hmm, I always manage to overcomplicate an easy question. thanks, I calculated my time to be 5.1 secs.

tal444

Your answer appears correct.

emailanmol

Always draw a diagram :-)
That shortens the problem in so many ways.

Your time is correct :-)
Now all that is left is to find the distance.