the equation is h=-0.4 d^2 + 2d.
h=height above water, d=horizontal distance.
Q1) how high above the water is the dolphin when it has travelled 2 m horizontally from its starting point?
A1) h=-0.4 (2)^2 + (2)
h=-1.6 + 2
h=0.4m
(im fairly sure that this is correct)
Q2) what horizontal distance does the dolphin cover when it first reaches a height of 25cm?
A2) 0.25=-0.4 d^2 + 2
0.25 + 0.4 = d^2 + d
0.65=d^2 +2
square root of 0.65=d + d
0.81= d+d
0.81/2=d
0.405m =d
Q3) what horizontal distance does the dolphin cover when it next reaches a height of 25cm? explain.
A3) i am unsure of this answer
Q4)what horizontal distance does the dolphin cover in one leap? (hint:what is the value of h when the dolphin has completed the leap?)
A4) i am unsure of this answer
Q5) can this dolphin reach a height of:
a)0.5 m
b)1 during a leap?
How can you work this out without actually solving the equation?
A5) i am unsure of this answer
Q6) find the greatest height that the dolphin reaches during its leap.
A6) i am unsure of this answer
CAN YOU PLEASE HELP ME WITH THIS INVESTIGATION!!!!!!!!!I NEED YOUR HELP SOON!!!!!!!!!!!!
The question is asking you for the horizontal distance to the point at which the dolphin reenters the water.
The dolphin enters/leaves the water when , so you need to find so that:
which may be factorised to give:
so the dolphin leaves the water when and re-enters the whater when
RonL