Cambridge 3U - Year 12
Ex 3G: Q6
 = 20cos30 = 20 \times \frac {\sqrt 3}{2} = 10\sqrt 3 ms^{-1})
 = 20 sin 30 = 10 ms^{-1} )
a)
 = \int 0 dt = C )
 = 10\sqrt 3 = C \implies \dot x(t) = 10\sqrt 3 )
 = \int 10\sqrt 3 dt = 10 \sqrt 3 t + D )
 = 0 = 0+D \implies D=0 \implies x(t) = 10\sqrt 3 t )
 = \int -10 dt = -10t + E )
 = 10 = 0 + E \implies \dot y(t) = 10 - 10t )
 = \int (10-10t)dt = 10t - 5t^2 + F)
 = 0 = 0-0+F \implies y(t) = 10t-5t^2)
b)
When height = -75m:
(t+3) = 0 \implies t = 5 s)
 = 10\sqrt 3 \times 5 = 50\sqrt 3 m )
c)
Greatest height occurs when vertical velocity = 0
 = 10-10t = 0 \implies t = 1 \implies y(1) = 10 \times 1 - 5 \times 1^2 = 5 m )
.: greatest height above river = 75m + 5m = 80m
d)
When pebble hits water:
 = 10\sqrt 3 ms^{-1} $ and $ \dot y(5) = 10-10 \times 5 = -40 ms^{-1})
^2 + (40)^2} = 10\sqrt {19} \approx 44 ms^{-1} )
It enters at angle given by:
e)
^2 = \frac {x}{\sqrt 3} - \frac {x^2}{60})
Ex 3E: Q10b(ii)
 = \frac {1}{x^2 + 6^2} \implies v^2 = 2\int \frac {1}{x^2+6^2}dx = \frac {1}{3}tan^{-1}(\frac {x}{6}) + C )
When t=0, v=0, x=0 ==> C = 0
 )
Since acceleration = 1/(x^2 + 36) > 0, and initial velocity = 0, velocity always > 0
For x = 6:
 = \frac {1}{3} \times \frac {\pi}{4}\implies v = +\sqrt{\frac {\pi}{12}} )
 $ \implies v^2 \to \frac {1}{3}tan^{-1} (\frac {\infty}{6}) = \frac {1}{3} \times \frac {\pi}{2} = \frac {\pi}{6})
Ex 3G: Q6
a)
b)
When height = -75m:
c)
Greatest height occurs when vertical velocity = 0
.: greatest height above river = 75m + 5m = 80m
d)
When pebble hits water:
It enters at angle given by:
e)
Ex 3E: Q10b(ii)
When t=0, v=0, x=0 ==> C = 0
Since acceleration = 1/(x^2 + 36) > 0, and initial velocity = 0, velocity always > 0
For x = 6:
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