A grasshopper leaps into the air from the edge of a cliff at a 50 degree angle. He reaches a maximum height 6.74 cm above the top of the cliff and travels a total horizontal distance of 1.06 m.

Question

1) How tall is the cliff?
2) What is the initial speed of the grasshopper?

Answer 1

y = a(x-h)^2 + k
y'(0) = tan 50° = 1.1917

k = 0.0674
a(1.06-h)^2 + k = 0
2a(-h) = 1.1917

y = -0.7782(x-0.7657)^2+0.0674
y = -0.7782x^2 + 1.1917x - 0.3887

Let the height y be given by

y = ax^2+bx+c
y(-b/2a) = 0.0674
y(1.06) = 0
y'(0) = 2ax+b = tan 50° = 1.1917

so, we have

b = 1.1917
a(-1.1917/(2a))^2 + 1.1917(-1.1917/(2a))+c = 0.0674
1.1236a+1.1917*1.06 + c = 0

-0.3550/a + c = 0.0674
1.1236a + c = -1.2632
a = -0.7783
c = -0.3887
y = -0.7783x^2 + 1.1917x - 0.3887

DANG IT!

I get the same solution, and while the vertex and x-intercept are correct, I have y<0 at x=0, which is not right.

Maybe you can find my error.