a) At the commencement of the ride, t = 0 (as the ride has only just begun), so substitute t = 0 into the formula (as Glockmeister said)
 - 45)^\circ + 16.5\\<br /> &= 15\sin (-45)^\circ + 16.5\\<br /> &= -15\sin (45)^\circ+ 16.5\\<br /> &= -15(\dfrac{\sqrt{2}}{2}) + 16.5\\<br /> &= 16.5 - \dfrac{15\sqrt{2}}{2}\\<br />\end{align*})
b) The access platform is 5 metres above the ground (this is what the question says), so you have to solve for t, i.e.:
^\circ + 16.5)
- Remember you only want the FIRST time that the platform is 5 metres above the ground, so just find the first positive answer.
c) This time, you have to solve the same equation as in part
b however, you need to consider all the answers in the domain
(As you only care about the first two minutes of motion). This question could also be answered with symmetry (i.e. finding the first two solutions then adding on the period until you go out of the domain).
d) Same as part
c except this time consider all
(as the ride goes for 6 minutes).
e) After 1 minute and 40 seconds, t = 100 s, so substitute t = 100 into the formula to get Isobel's height above ground.
f) For this one, think about it like this, Hamish and Isobel's height will be similar, except Hamish's will be horizontally translated so that at t = 0, h = 1.5. After finding an equation for Hamish's height, substitute t = 100 and solve for the height.
Let Hamish's Height = B
^\circ + 16.5)
where c is some horizontal translation.
now we know that
 = 1.5)
, so:
 - 45 + c)^\circ + 16.5\\<br />-15 &= 15\sin (c - 45)^\circ\\<br />-1 &= \sin(c - 45)^\circ\\<br /> c &= 315^\circ<br />\end{align*}<br />)
substituting this back into B, we get:
^\circ +16.5\\)
Then substituting t = 100 will get your answer
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