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
2 answers
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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
If you need 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
If you need to find the maximum height then:
The coordinate of the point at which the quadratic function has an extreme:
t = - b / 2 a
in this case:
t = - 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
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
If you need 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
If you need to find the maximum height then:
The coordinate of the point at which the quadratic function has an extreme:
t = - b / 2 a
in this case:
t = - 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