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
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
... and ... ?
To determine the time it takes for the arrow to hit the ground, we need to find when the height is equal to zero.
Given:
h = -5t^2 + 15t + 20
Set h = 0:
0 = -5t^2 + 15t + 20
This is a quadratic equation in the form of ax^2 + bx + c = 0, with a = -5, b = 15, and c = 20.
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
Substituting the values, we get:
t = (-(15) ± √((15)^2 - 4(-5)(20)))/(2(-5))
Simplifying:
t = (-15 ± √(225 + 400))/( -10)
t = (-15 ± √(625))/(-10)
t = (-15 ± 25)/(-10)
So, the two possible solutions for t are:
t = (-15 + 25)/(-10) or t = (-15 - 25)/(-10)
Simplifying:
t = 10/-10 or t = -40/-10
t = -1 or t = 4
Therefore, the arrow hits the ground at t = -1 second or t = 4 seconds after it was released.
To find the time it takes for the arrow to hit the ground, we need to determine when the height of the arrow, h, is equal to zero.
Given the equation h = -5t^2 + 15t + 20, we can set it equal to zero:
0 = -5t^2 + 15t + 20
Now, we can solve this quadratic equation to find the values of t.
First, let's reformat the equation to standard quadratic form by multiplying both sides by -1 to simplify it:
0 = 5t^2 - 15t - 20
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is a bit more complicated, so let's use the quadratic formula:
The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation of 5t^2 - 15t - 20, we have a = 5, b = -15, and c = -20.
Now let's plug these values into the quadratic formula:
t = (15 ± √((-15)^2 - 4*5*(-20))) / (2*5)
Simplifying further:
t = (15 ± √(225 + 400)) / 10
t = (15 ± √625) / 10
t = (15 ± 25) / 10
This gives us two possible values for t:
t1 = (15 + 25) / 10 = 40 / 10 = 4 seconds
t2 = (15 - 25) / 10 = -10 / 10 = -1 second
Since time cannot be negative in this context, we discard t2 = -1 second. Therefore, the arrow takes approximately 4 seconds to hit the ground.