We can model the equation of the height of the rocket as ∩-shape curve as shown below
Part A:
The time when the height is 720 feet
, rearrange to make one side is zero
, divide each term by 16
, factorise to give
and
So the rocket reaches the height of 720 feet twice; when t=5 and t=9
Part B:
We will need to find the values of t when the rocket on the ground. The first value of t will be zero as this will be when t=0. We can find the other value of t by equating the function by 0
and
and
So the time interval when the rocket was launched and when it hits the ground is 14-0 = 14 seconds
(5/50) x 100 = 10%
(3/20) x 100 = 15%
So Rachel has the biggest discount
Answer:
4. 118 kg
w = 450 j , d = 3.8 m
w = mgh ⇒mg = w / h ⇒m = 450/3.8 = 118 kg
5. 9806.65 Joules
6. 25 J
P=2(w+h)
remember you can do anythig to an equaotn as lng as you do it to both sides
divide both sides by 2
P/2=w+h
subtract w
(P/2)-w=h
I'm sorry but what's the full question?