The height in metres of a stone thrown by a student t seconds after it is thrown into the air is given by H(t)=-5t^2+30t+2a) Explain why it is -5^2...

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