A basketball is tossed from the top of a 3m wall. The path of the basketball is defined by the relation �y=-x^2+2x+3 where x represent the horizontal distance travelled, in metres, and y �represents the height, in metres, above the ground.

Question

A) How far has the basketball travelled horizontally when it lands on the
ground?

3m. Because when I solved it, the roots were 0=*x+1)(x-3) and x-3=0 so x would be 3m.

B) What is the highest height the ball reaches?
Vertex is (1,4). Highest height would be 4m.

C) For how many seconds is the ball higher than 3.5 m?

How would I do this? Do I just plug 3.5 as y and solve? But that doesn't seem right.

Answer 1

a) there are two solutions, one is negative (ignore that).
0=-x^2+2x+3
0= x^2-2x-3=(x-3)(x+1)
so indeed, x=3
B) vertix is indeed 1,4
C)time in air. You need something else. It would be nice to know its horiztal velocity...