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?
Do you just want just the answer or wnat to know how to find it?
Can you show me how to find it please with the ans
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
y do u put h as zero?
im confused.
can yu show us what yu plugged in for 4)
this ball will reach a maximum heigh (h) then it returns back the same height (-h) so it's total value = 0
oohhh i get it
BTW..this will be from the position of projection , not the surface of the cannon
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.
can yu put the equation where yu hav to isolste it so we kno wht to rite
or put in our calculator
isolate "t"
actually I can't find anything confusing here is the problem with expanding of the brackets ?!
no i jus dun understand how to solve 4)
like how do u isolate "t" to find the time?
...
did you understand why h = 0 ?!!!
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
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]
thank you :D charmi - u get it now? :P
yupp
I'm glad you understood :)
ty
Join our real-time social learning platform and learn together with your friends!