Question
The height h (in meters) of a cannonball t seconds after it is fired into the air is given by
h=-4t^2+16t+9.
a) What is the initial height of the cannonball? What feature does this correspond to on the graph?
b) Find the height of the cannonball after 1 second.
c) What is the maximum height of the cannonball? When does it reach the maximum height?
d)How long does it take for the cannonball to fall back to earth?
h=-4t^2+16t+9.
a) What is the initial height of the cannonball? What feature does this correspond to on the graph?
b) Find the height of the cannonball after 1 second.
c) What is the maximum height of the cannonball? When does it reach the maximum height?
d)How long does it take for the cannonball to fall back to earth?
Answers
Damon
a) Hey, you know what h is when t = zero!
b ) h = -4(1) + 16(1) + 9
= 21
c) If you do not know any calculus, which I assume you do not or you would not be asking, then you must complete the square to find the vertex of the parabola
t^2 - 4 t = -h/4 + 9/4
t^2 - 4 t + 4 = -h/4 + 9/4 + 16/4
(t - 2)^2 = -(1/4) (h - 25)
so
max height = 25 at t = 2
d ) 0 =-4t^2+16t+9
or
t^2 - 4 t - 9/4 = 0
t = [ 4 +/- sqrt(16 -81/4) ] /2
no real solution
The reason is that your original equation is in error
h=-4t^2+16t+9.
It should be assuming feet and seconds and g = -32 ft/s^2
h = (1/2) g t^2 + Vi t + Hi
h = -16 t^2 + Vi t + Hi
I suspect a typo
b ) h = -4(1) + 16(1) + 9
= 21
c) If you do not know any calculus, which I assume you do not or you would not be asking, then you must complete the square to find the vertex of the parabola
t^2 - 4 t = -h/4 + 9/4
t^2 - 4 t + 4 = -h/4 + 9/4 + 16/4
(t - 2)^2 = -(1/4) (h - 25)
so
max height = 25 at t = 2
d ) 0 =-4t^2+16t+9
or
t^2 - 4 t - 9/4 = 0
t = [ 4 +/- sqrt(16 -81/4) ] /2
no real solution
The reason is that your original equation is in error
h=-4t^2+16t+9.
It should be assuming feet and seconds and g = -32 ft/s^2
h = (1/2) g t^2 + Vi t + Hi
h = -16 t^2 + Vi t + Hi
I suspect a typo
Reiny
I agree with Damon that your gravitational constant is off.
Since you are using metric , your equation should be
h = -4.905t^2 + 16t + 9
let's ballpark it to
h = -5t^2 + 16 + 9
a) .....
b) sub in t = 1 , h = -5+16+9 = 20 m
c) for t of the vertex ...
t =-b/(2a) = -16/-10 = 1.6
h = -5(1.6^2) + 16(1.6) = 9 = 21.8
Maximimum height of 21.8 m after 1.6 seconds
d) -5t^2 + 16t + 9 = 0
5t^2 - 16t - 9 = 0
t = (16 ± √436)/10
= 3.69 or a negative
it will hit the ground after 3.69 seconds
Since you are using metric , your equation should be
h = -4.905t^2 + 16t + 9
let's ballpark it to
h = -5t^2 + 16 + 9
a) .....
b) sub in t = 1 , h = -5+16+9 = 20 m
c) for t of the vertex ...
t =-b/(2a) = -16/-10 = 1.6
h = -5(1.6^2) + 16(1.6) = 9 = 21.8
Maximimum height of 21.8 m after 1.6 seconds
d) -5t^2 + 16t + 9 = 0
5t^2 - 16t - 9 = 0
t = (16 ± √436)/10
= 3.69 or a negative
it will hit the ground after 3.69 seconds
Abol
d) use quadratic formula
-b +- square root of b^2 - 4 times a times c devided by 2a
= -16+-square root 400 devided by -8
= 4.5 secounds.
-b +- square root of b^2 - 4 times a times c devided by 2a
= -16+-square root 400 devided by -8
= 4.5 secounds.