H(1) = 20 – 4.9(1)^2

H(1) = 20 – 4.9(1)

H(1) = 20 – 4.9

H(1) = 15.1

H(1.1) = 20 – 4.9(1.1)^2

H(1.1) = 20 – 4.9(1.21)

H(1.1) = 20 – 5.929

H(1.1) = 14.071

H(1.2) = 20 – 4.9(1.2)^2

H(1.2) = 20 – 4.9(1.44)

H(1.2) = 20 – 7.056

H(1.2) = 12.944

B.) This part of the problem gives you the values for “H(x)” – which is essentially the y value instead of the value. Again, use substitution and then solve for the time (the x) instead.

15 = 20 – 4.9x^2

4.9x^2 = 5

x^2 = 1.02041

x = 1.01053

10 = 20 – 4.9x^2

4.9x^2 = 10

x^2 = 2.04082

x = 8/7

5 = 20 – 4.9x^2

4.9x^2 = 15

x^2 = 3.06122

x = 1.74964

C.) This part is similar to part ‘b,’ but you have to decide what height you want to substitute into the equation. Because you are being asked when the rock strikes the ground, you’re looking for the rock’s height on the ground. This, of course, would be zero.

0 = 20 – 4.9x^2

4.9x^2 = 20

x^2 = 4.08163

x = 2.02031

The rock strikes the ground after about 2.02031 seconds.

ex- H=20-4.9(1)^2

ex.H=20-4.9(1.1)^2

also what doe ^ mean?

if i know what ^ is i could probley help you with the rest because i just did these ind of questions a few months a ago and i still remember how to do it