H(t) = -5t^2 +30t + 2

a) The factor for t^2 is negative because the rock was thrown against the gravity. If the rock was falling toward gravity, then the factor will be positive.

b)To reach maximum height. We need to find the maximum value of H(t):

First we need to determine first derivative's zeros.

H(t) = -5t^2 + 30 t + 2

==> H'(t) = -10t + 30 = 0

==> t= 3

B) Then the height will be at maximum point when the time t= 3

C) To find the maximum height we will substitute with t=3.

Then the maximum hight is:

H(3) = -5*3^2 + 30*3 + 2

= -45 + 90 + 2 = 47

Then the maximum height is 47 .

d) The time before it hits the ground. This means we need the time when the height is 0. (hit the ground)

So, we will substitute with h= 0 and determine t:

Note that there will be two values for t, the first when first we through the rock, and the second will be when it hit the ground.

H(t) = -5t^2 + 30t + 2

==>-5t^2 + 30 t + 2 = 0

==> t1= [-30 + sqrt(900-4*-5+2)]/2*-5

= [-30 + sqrt(940)]/-10

= (-30+ 30.7)/-10

= (-0.7/-10 = 0.07

==> t1= 0.07 ( thisis te initial time when the roch first thrown.

==> t2= (-30 -30.7)/10 = -60.7/-10 = 6.07

Then the time needed for the roch to hit the ground is:

t= 6.07 seconds.

e) The height where the och thrown is:

H(t1) = H(0.07) = 4.0755