anonymous
  • anonymous
A cannonball is fired into the air. It's height, h, in metres after t seconds is given by the relation: h=-4.6 (t-7.5)^2+285 a) what is the maximum height reached by the cannon and what time is this reached. b) what was the initial height of the cannon? c) what is the height of the cannon after 10.5 seconds? d) how long is the cannon in the air?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmate
  • mathmate
Do you just want just the answer or wnat to know how to find it?
anonymous
  • anonymous
Can you show me how to find it please with the ans
y2o2
  • y2o2
h = -4.6t²+69 t+26.25 the vertex of this function is the the time of maximum height time of maximum height = -b/2a = 7.5 maximum height = -4.6(7.5)²+69(7.5)+26.25 = 285 meters 2) initial height (t = 0) = 26.25 meters 3) after 10.5 sec. (t = 10) = 243.6 meters 4) time in air (h = 0) t = 15.37

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
y do u put h as zero?
anonymous
  • anonymous
im confused.
anonymous
  • anonymous
can yu show us what yu plugged in for 4)
y2o2
  • y2o2
this ball will reach a maximum heigh (h) then it returns back the same height (-h) so it's total value = 0
anonymous
  • anonymous
oohhh i get it
y2o2
  • y2o2
BTW..this will be from the position of projection , not the surface of the cannon
mathmate
  • mathmate
The answer to part a is already given in the way the formula is expressed: h=-4.6(t-7.5)^2+285 is in the vertex form. It shows that the vertex (highest or lowest point) is at the position (7.5,285), or a height of 285 after 7.5 seconds.
anonymous
  • anonymous
can yu put the equation where yu hav to isolste it so we kno wht to rite
anonymous
  • anonymous
or put in our calculator
anonymous
  • anonymous
isolate "t"
y2o2
  • y2o2
actually I can't find anything confusing here is the problem with expanding of the brackets ?!
anonymous
  • anonymous
no i jus dun understand how to solve 4)
anonymous
  • anonymous
like how do u isolate "t" to find the time?
anonymous
  • anonymous
...
y2o2
  • y2o2
did you understand why h = 0 ?!!!
anonymous
  • anonymous
yeah we understand that, we just dont get how to put the "t" on one side of the equation, and the rest on the other side of the equation
y2o2
  • y2o2
well, as h = 0 then h = -4.6 (t-7.5)²+285 = 0 -4.6 (t-7.5)² = -285 [divide by -4.6] (t-7.5)² = 61.95 [approx.] t-7 = 8 or t-7 = -8 [by taking +ve and -ve square root] t = 15 or t = -1 [refused answer bec. t is never -ve]
anonymous
  • anonymous
thank you :D charmi - u get it now? :P
anonymous
  • anonymous
yupp
y2o2
  • y2o2
I'm glad you understood :)
anonymous
  • anonymous
ty

Looking for something else?

Not the answer you are looking for? Search for more explanations.