Can someone please help me with Maths Methods homeworkrk?

Question

Can anyone please help with the workings out for this question

A ball is thrown vertically up so that its height above the ground, h metres, at any time, t seconds, after leaving the thrower’s hand is given by the function h(t) = t − t^2 + 2.

a) Find the height of the ball as it leaves the thrower’s hand.

b) Find when and where the ball reaches its greatest height.

c) Find when the ball returns to the same level that it left the thrower’s hand.

d) If the ball isn’t hit, find when the ball hits the ground to the nearest

thousandth of a second.

e) Hence state the domain and range of h(t).

f) Sketch the graph of h versus t.

Any help would be greatly appreciated

Answer 1

a) let t=0  thus h(0)=2

b)set h'(t) to zero and solve

-2t+1=0

t=1/2

c) set h(t)=2 solve

thus 0=t(1-t)

t=0 and t=1

d) set h(t)=0

thus 0=t^2-t-2

         = (t-2)(t+1)

t=2

since is time use logic and t=-1 is irrelavent

e) domain: is values of time so goes from 0 to 2.  after 2 seconds the function no longer makes any since it will lead to negative heights

range is values of hieghts so it goes from 0 to the point of its greatest height which was at t=1/2 so h(1/2)=1/2-1/4+2=2 and 1/4 or 2.25 or 9/4

   range:

f) i cant graph here so just use the points we found and graph them

Answer 2

a)  when leaves thrower's hand  t = 0

    subsitute in  eq

       we get h = 2

b)greatest height is reached when dh/dt = 0

   ie  dh  / dt = 1-2t   = 0  t  =1/2

    t  = 1/2 sec

    sub in  eq  = 1/2 - 1/4 +2  =  9/4

c) when ball reaches same level h =2

       t-t^2 +2  = 2

      t- t^2  =0

      t(1-t)  = 0

      t = 0 or   t = 1

     t  = 0 when he throws   t  =1 it reaches again

d) t - t^2 +2  = 0 ( when it hits ground h = 0)

t^2-t-2  =0 solving for t

t =  (1 + √1+8)/2  ( t cannot be -ve)

t = 2sec

e)

normally The domain of  function is the set of all real numbers. The range is the set of values that f(x) takes as x varies

in this case  domain is ( 0 , infinity)   range  (0,9/4)

f)  to plot the graph give t diff values between t = 0  to t = 2 in steps of 0.25

and plot

Answer 3

a) It will leave the thrower's hand at t=0. Therefore,

    h=0 - 0^2 + 2

    h=2

b) This is trial and error:

    

    assume t= 1

    h= 1 - 1^2 + 2

    h=2

    assume t=0.5

    h= 0.5 - 0.5^2 + 2

    h=2.25

    

    assume t= 2

    h=2 - 2^2 +2

    h=0

    Therefore, it will reach the highest point 2.25h at 0.5t.

c)  0.5t to reach it's highest point, so therefore it will take the same t to go back to it's initial height.  

    

    0.5t x 2= 1t (or just t)

not sure of the rest. Hope this helped

    

Answer 4

a) The height of the ball as it leaves the thrower’s hand is at t=0,

So h(0) = 2

b) Greatest height is when dh /dt = 0

                           => 1 - 2t = 0

                           => t = 1/2 seconds

                       hmax =h (t =1/2 seconds) = 1/2 - 1/4 + 2 = 2.25m

c) When ball hits grounds, h(t) = 0

So t -t^2 + 2 =0

=> t^2 - t - 2 = 0

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

=> (t +1)( t- 2) =0

So t = 2 seconds

d) domain and range

http://www.analyzemath.com/DomainRange/D...