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# M2 Projectile questions watch

1. Number 1 and 6 from the question sheet

I've done part a, and got an answer of 1.34. Not sure if that's right, but I posted my working in an attachment. In part B, I don't know where start.

And for 6, do you use sin (2theta) = 2 sin(theta) cos(theta), or get some sort of quadratic.

Attached Images

2. for 1b)

s = ut + ½at²

a = 0, t = 1.34, u = 13cos27

For 6,

this is gonna hard to type out but whatever :

Vertical
s = ut + ½at²
0 = (25sinθ)t + ½(-9.8)t²
4.9t² - (25sinθ)t = 0

Horizontal
s = ut
60 = (25cosθ)t
t = 12/(5cosθ)

substitute your t into the equation for the vertical

4.9t² - (25sinθ)t = 0
4.9(12/(5cosθ))² - (25sinθ)(12/(5cosθ)) = 0
28.224 sec²θ - 60tanθ = 0
28.224 (1 + tan²θ) - 60tanθ = 0
28.224tan²θ - 60tanθ + 28.224 = 0

solve for tanθ

tan θ = 1.423 and tanθ = 0.70623

θ = 54.9 and 35.1 respectively

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Updated: September 26, 2010
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