Question

Let h=v0^2/4.9 sin(theta) cos(theta) model the horizontal distance in meters traveled by a projectile. If the initial velocity is 44 meters/second, which equation would you use to find the angle needed to travel 150 meters?

A. 8.98sin(2 theta) = 150

B. 395.10sin(2 theta) = 150

C. 150sin(2 theta) = 150

D. 197.55sin(2 theta) = 150

@terenzreignz what do u think? this one, i don't get.. like at all :/ haha :P

Answer 1

i'm kinda leaning towards B... but i'm not too sure on this whole process :/

Answer 2

I see... hang on a sec.

Answer 3

kk :)

Answer 4

Okay... you clearly see that v0 = 44, right?

Answer 5

yes :) do i need to square that? :/

Answer 6

like 44^2 ? :/ not sure if that's what I'm supposed to do?

Answer 7

Okay, tell you what, let's put this range formula in its more traditional form...
\[\large \frac{v_o^2}{4.9}\sin(\theta)\cos(\theta)\]

Answer 8

kk :)

Answer 9

So, we can multiply and divide by 2... to keep things the same
\[\large \frac{v_o^2}{4.9}\cdot\frac12\cdot2\sin(\theta)\cos(\theta)\]

Answer 10

Giving us
\[\large \frac{v_o^2}{9.8}\cdot2\sin(\theta)\cos(\theta)\]

Answer 11

seriously, latex rocks! :)

haha oaky! :D

Answer 12

\[\large \frac{v_o^2}{9.8}\cdot\color{red}{2\sin(\theta)\cos(\theta)}\]
This bit here is the very definition of \(\large \sin(2\theta)\)
So we get
\[\large \frac{v_o^2}{9.8}\cdot\sin(2\theta)\]
Now plug in your \(\large v_o = 44 \)

Answer 13

okay... ermm sooo 44^2/9.8 x sin(2 theta) ?

Answer 14

yup...

Answer 15

so 1936/9.8=197.55 right?

Answer 16

yes.

Answer 17

so the answer is D. 197.55sin(2 theta) = 150 ?? :D

Answer 18

What does your heart tell you?

Answer 19

that it's D :) hehe

Answer 20

hehe thank youu :)