The general form of this quadratic equation is:
h = a t² + b t + c
In this case:
h = - 5 t² + 15 t + 20
The coefficients of this quadratic equation are:
a = - 5 , b = 15 , c = 20
The initial height of the arrow is:
h ( t = 0 ) = h(0) = a ∙ 0² + 15 ∙ 0 + 20 = 20
To find when an arrow touches the ground solve the equation:
h = - 5 t² + 15 t + 20 = 0
The solutions are:
t = - 1 and t = 4
Time can not be negative, so:
t = 4
Maximum height.
The coordinate of the point at which the quadratic function has an extreme:
t = - b / 2 a
in this case:
t = - b / 2 a = - 15 / 2 ∙ ( - 5 ) = - 15 / - 10 =
15 /10 = 5 ∙ 3 / 5 ∙ 2 = 3 / 2 = 1.5
h(max)= h( x =1.5 ) = h ( 1.5 ) = - 5 ∙ 1.5² + 15 ∙ 1.5 + 20 = 31.25
Over what time interval is the arrow above 26:
Solve the equation:
- 5 t² + 15 t + 20 = 26
The solutions are:
t = 0.4753 and t = 2.5247
This means the arrow will have a height of 26 twice.
For t = 0.4753 when climbing
and
for t = 2.5247 when falling towards the ground.
an arrow is shot from a castle tower. its height above the ground is given by the equation h = -5t^2 + 15t + 20 where h is the height of the arrow, in meters, and t is the time, in seconds, after the arrow was released
what is the initial height of the arrow?
what time does the arrow hit the ground?
what is the maximum height of the arrow?
over what time interval is the arrow above 26m?
2 answers
My little typo.
Don't write:
h ( t = 0 ) = h(0) = a ∙ 0² + 15 ∙ 0 + 20 = 20
Write:
The initial height of the arrow is:
h ( t = 0 ) = h(0) = ( - 5 ) ∙ 0² + 15 ∙ 0 + 20 = 20
Don't write:
h ( t = 0 ) = h(0) = a ∙ 0² + 15 ∙ 0 + 20 = 20
Write:
The initial height of the arrow is:
h ( t = 0 ) = h(0) = ( - 5 ) ∙ 0² + 15 ∙ 0 + 20 = 20