Answer:
0m
Step-by-step explanation:
Before the ball is thrown, it is at the ground level. At the ground level, the height is zero. Substitute h = 0 into the equation;
h(x)=-x^2+10x-16
0=-x^2+10x-16
x^2-10x+16 = 0
Factorize
x^2-8x-2x+16 = 0
x(x-8) - 2 (x-8) = 0
x-2 = 0 and x - 8 = 0
x = 2 and 8
At x = 2
h(2) = -2^2 + 10(2) - 16
h(2) = -4 + 4
h(2) = 0
Hence the height of the ball at the time it is thrown is 0m
