+147 1. Aurora hit a baseball with an initial velocity of 70 feet per second at an angle of 30° with the horizontal. The ball hit her bat when the ball was 3 feet above the ground.
(a) No one interferes with the ball. How long does it take the ball to hit the ground? Round your answer to the nearest hundredth of a second. Show all your work.
(b) How far did the ball travel horizontally? Use your answer from Part (a) in your calculations. Round your answer to the nearest tenth of a foot. Show all your work.
+118687 1) Find or state all the initial conditions.
What is the height?
the vertical speed and the horizontal speed?
and the acceleration in vertical and horizontal directions?
To find the initial vertical and horizontal speeds.
Draw a right-angled triangle. The right angle and 30 degrees will be at the base
The hypotenuse will be 70feet/sec
Use trig to find the initial vertical and horizontal speeds.
When you have done that get back with the worked answers (or explain your confusion)
+118687 Are they the worked answers to the question that I actually asked?
I don't know if they are the final answer or not. If you know how to do it then you don't need my help.
+147 I didn't really understand your explanation so I reposted the questions.
I should have rephrased my question. I wanted to know if what I did was right. Me and my friend were working on the same question and we both got different answers and we have a test in a couple of days so we were trying to figure out who was right.
+118687 If you did not understand my response then you cannot do the question.
Reposting was downright rude.
You have not shown us anything at all that you have done.
You have presented 2 numbers that could have come from anywhere.
+147 I'm really sorry! I did not mean to be rude
a)
y = -16t^2+(vsinθ)t + h
y = -16t^2 + (70sin30)t+3
0= -16t^2 + (70*1/2)t + 3
I solved for t through the quadratic formula and got t = 2.27 seconds
b) I plugged in t = 2.27 seconds into x = (vcosθ)t
x = (70cos30)(2.27)
= 35 sqrt(3)(2.27)
x = 137.6 feet
+118687 ok, perhaps you have a major problem with reading and interpreting questions. Although I doubt you seriously read my response.
I did not ask you how to get to the final answer.
I asked you to tell me the initial conditions.
I asked you for the initial vertical speed, the initial horizontal speed, the initial acceleration, the initial vertical and horizontal displacement.
You still have not given me those. Nor have you discussed them in any way.
You cannot seriously understand your original question if you cannot answer my questions.
+147 Oh! You wanted me to find the initial condition, height, vertical speed, horizontal speed, and accelerations.
I thought you wanted me to use these values to find my answer. I really wasn't trying to ignore your questions. I'm sorry if it came out that way.
Initial speed: v = 70 feet per second
height: h = 3 feet
vertical speed: 35
horizontal speed: 60.62
Once again, I am extremely sorry. It was a big misunderstanding.
+118687 ok Thanks for answering.
Set these questions out very logically and methodically. It will help you think each one through properly.
Start with one or more pics to illustrate what is happening.
Here is my initial condition pic with working. A little more elaborate than what you need.
v is vertical velocity
u is horizontal velocity
initial height = +3feet
acceleration is gravity and it stays the same all the time at 32feet/sec^2 but it is pulling down so it is negative.
And it only affects vertically movement.
\(\frac{dv}{dx} =-32\)
Wind and air resistance is ignored for you so there is no horizontal forces that you need to worry about so what is horizontal acceleration?
+118687 No, this is a calculus question.
I understand now why you appeared to ignore my initial answer. You had no idea what I was talking about. 
+147 Well, I mean I just complete Pre-Calculus 2 days ago so probably.
I really didn't mean to ignore your questions.
I got the answer right though!!! My friend and I asked my teacher how to do it and she said that I was correct!!! YAY!
Thank you for your time!!! I really appreciate it!
+118687 ok, you have found your answers, it might be correct. I haven't checked.
You understand none of it.
That is not your fault, I am not blaming you, it is just an observation.
Maybe you will understand more as the topic progresses.
I hope that you do and I wish you well.
+147 We were thought to put this in a formula x = (vcos(theta))t and y = -16t^2+(vsin(theta))t+h so, that's probably why I didn't understand the acceleration part!
Thank you for the wishes!
+118687 You are learning it by physics formulas rather than by calculus. That is ok. I think that is more common in the US than here.
(I assume you are in the US, apologies if I am wrong)
Still, it would be good for you to look closely and understand my diagram.
Then you will understand where the initial vertical velocity formula comes from. If you understand it then you don't have to memorize it.
For me the less formulas I have to memorize, the better.
+147 Ok, that makes sense!
Yeah, you were right, I am in the US!
I understand the diagram and I will try to understand the acceleration part.
But, Thank you so much for helping me!
And, I am so sorry for sounding rude!
