Applications of Differentiation?

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) =(8/3) t − (8/9)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

h(t) =(8/3) t − (8/9)t^2 + 2.

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

h=2

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

dv/dt=(-16/9)t+8/3

when v=0....0=(-16/9)t+8/3

t=(8/3)/(16/9)=3/2

while the hieght=(8/3) t − (8/9)t^2 + 2.

h(3/2)=(8/3)(3/2)-(8/9)(9/4)+2

hieght=4-2+2

h=4

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

ofcourse the total time is twice the time of its max. hieght

t=(3/2)2=3

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

thousandth of a second.

let h=0

0= (8/3) t − (8/9)t^2 + 2

0=  t^2 -(1/3)t- 3/4

t^2 -(1/3)t=3/4

(t-1/6)=sqrt(3/4+1/36)

t=[sqrt(3/4+1/36)]+1/6 just use your calculator to solve

thats all i can help

......

Answer 2

a) When it leaves the thrower's hand t = 0

      

       Therefore, height of the ball = h(0) = 2.

b) To find when the ball reaches the greatest height you will have to find the point of max. value of h(t).

Now,

       h'(t) = 8/3 - 16t/9   (h'(t) & h"(t) are 1st & 2nd. order derivative of

                                    h(t) with respect to t)

    Putting h'(t) = 0 we get t = 3/2

    h"(t) = -16/9 < 0, therefore t=3/2 is point of max. value.

Therefore the ball reaches the highest value at t = 3/2 s and at that time it's height is 8/3 * 3/2 - 8/9 * 9/4 + 2 = 4 metres.

c)Wheh the ball left the thrower's hand it's heigt was 2m.

   therefore putting h(t) = 2 and solving for t we get,

                        2 = (8/3)t - (8/9)t^2 + 2

                     => t = 0 or 3 s.

Therefore the ball returns to the thrower's hand at t= 3s.

d) When the ball hits the ground it's height is 0m.

    Therefore putting h(t) = 0 and solving for t we get

                        t = (3 + 3 root2)/2 or (3 - 3root2)/2 <---it is -'ve

so t = (3 + 3root2)/2 = 3.621s

e) domain of h(t) is [0, (3 + 3root2)/2] and it's range is [0, 4].

f) Sorry, I can't draw graph here.

hah! it's over.