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?
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.
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
To find the initial height of the arrow, we need to look at the equation given. The initial height is the height of the arrow when t = 0. So we substitute t = 0 into the equation:
h = -5(0)^2 + 15(0) + 20
h = 0 + 0 + 20
h = 20
Therefore, the initial height of the arrow is 20 meters.
To find the time when the arrow hits the ground, we need to find the value of t when the height, h, is 0. So we set h = 0 in the equation and solve for t:
0 = -5t^2 + 15t + 20
This is a quadratic equation that can be solved either by factoring or by using the quadratic formula. Let's use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
In this equation, a = -5, b = 15, and c = 20. Substituting these values into the equation, we get:
t = (-15 ± sqrt(15^2 - 4(-5)(20))) / 2(-5)
t = (-15 ± sqrt(225 + 400)) / -10
t = (-15 ± sqrt(625)) / -10
t = (-15 ± 25) / -10
There are two possible solutions:
1. t = (-15 + 25) / -10 = 10 / -10 = -1
2. t = (-15 - 25) / -10 = -40 / -10 = 4
Since time cannot be negative in this context, we disregard t = -1. Therefore, the arrow hits the ground at t = 4 seconds.
To find the maximum height of the arrow, we can use the vertex formula. The vertex of a parabola is the highest or lowest point on the curve. The formula is given by:
t = -b / (2a)
In this case, a = -5 and b = 15. Substituting these values into the formula, we get:
t = -15 / (2(-5))
t = -15 / -10
t = 1.5
Now we substitute t = 1.5 into the equation to find the maximum height:
h = -5(1.5)^2 + 15(1.5) + 20
h = -11.25 + 22.5 + 20
h = 31.25
Therefore, the maximum height of the arrow is 31.25 meters.
To find the time interval during which the arrow is above 26 meters, we need to find the values of t for which h is greater than 26. So we set h > 26 in the equation and solve for t:
-5t^2 + 15t + 20 > 26
Rearranging the equation, we get:
-5t^2 + 15t - 6 > 0
Solving this quadratic inequality requires factoring or using the quadratic formula. Let's use factoring:
-5t^2 + 15t - 6 = 0
The factored form is:
(-5t + 2)(t - 3) > 0
Now we set each factor equal to zero and solve for t:
-5t + 2 = 0
t = 2/5
t - 3 = 0
t = 3
These are the critical points. We plot them on a number line and check the inequality between each interval:
(-Infinity, 2/5) -> -5t + 2 < 0 and t - 3 < 0 --> Both factors negative, so the inequality is not satisfied.
(2/5, 3) -> -5t + 2 < 0 and t - 3 > 0 --> First factor negative, second factor positive, so the inequality is satisfied.
(3, Infinity) -> -5t + 2 > 0 and t - 3 > 0 --> Both factors positive, so the inequality is satisfied.
Therefore, the arrow is above 26 meters during the time interval t ∈ (2/5, 3).