The formula h=30+40t-5t^2 can be used to find the height h, in meters, of an object shot upward from the top of a building 30 m above the ground at the initial speed of 40 mps, after t seconds. How much higher above the ground is the object after 4 seconds than 2?

20m

Step-by-step explanation:

Given the formula that modelled the height given as;

h=30+40t-5t^2

At t = 4s

h=30+40(4)-5(4)^2

h = 30+160-80

h = 160-60

h = 110m

At t = 2s

h=30+40(2)-5(2)^2

h=30+80-20

h = 110-20

h= 90m

Difference in height = 110-90 = 20m

Hence the height is 20m higher above the ground at 4s than at 2s

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)