# Need help on projectile

#### ataglance05

need help on projectile!!

1. The problem statement, all variables and given/known data
A projectile is fired at an angle of 38 degrees with respect to the x axis and with an initial velocity in the x direction of 3.3 m/sec. (gravity=10m/sec^2)

a)whats the initial velocity of the projectile?

2. Relevant equations
a) i think Vy=Vi(sin38 degrees)
and a^2+b^2=c^2

3. The attempt at a solution
I first want to get the Vy by using: Vy=Vi(sin theta)
so, Vy=3.3m/sec(sin 38 degrees)
Vy=2.03 m/sec

now, (2.03)^2 + (3.3)^2= r^2
so r or the initial velocity of the projectile =3.87 m/s ????????

#### OlderDan

Homework Helper
Your reasoning is circular and incorrect. First you are using Vi to find Vy, and then you are using Vy and Vx to find what you called r, which is in fact Vi. You are not getting r = Vi because you did not know Vi when you used it to find Vy. What you are given in the problem is Vx and a launch angle. How do you find Vi from the given information?

#### ataglance05

let me try again, with the right variables and numbers: Viy=Vi/(tan theta)
Viy=3.3 m/sec / (tan 52 degrees)
Viy=2.58 m/sec

so then i do the phythagorem (sp?) theorm:
Vix+Viy=Vi
(3.3^2)+(2.58^2)=Vi
4.19 m/sec=Vi !!!!!!!!?!??!?!

i believe thats correct

#### cristo

Staff Emeritus
let me try again, with the right variables and numbers: Viy=Vi/(tan theta)
Viy=3.3 m/sec / (tan 52 degrees)
Viy=2.58 m/sec

so then i do the phythagorem (sp?) theorm:
Vix+Viy=Vi
(3.3^2)+(2.58^2)=Vi

4.19 m/sec=Vi
i believe thats correct
Your answer is correct. However the lines in bold should read Vi2=Vix2+Viy2
Vi=sqrt{(3.3^2)+(2.58^2)}.

The Theorem to which you are referring is the Pythagorean Theorem (named after Pythagoras).

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