HOW DO YOU DO PART TWO?
The performer stands on a tightrope 3m above ground and theows a diabolo upwards. In each throw , the height (ym) of the diabolo above the ground after s seconds is given by y=-s^2+3s+4
1) Find maximum height that the diabolo can reach
My answer: 25/4 metres
(The answer key proved it to be correct)
2) unfortunately he loses his grip and the diabolo falls to the ground directly. Find the time that the diabolo takes to reach the ground after it is thrown
Thankyou!!!
as usual, the vertex is at s = -b/2a = 3/2
y(3/2) = 25/4
so far, so good. So, when does it hit the ground? When y=0, naturally. So,
-s^2+3s+4 = 0
(s+1)(4-s) = 0
so, it hits the ground after 4 seconds.