We are given an equation for the height of the ball, where θ is the angle between the ground and the path of the ball, v_0 is its initial velocity in meters per second and g is the acceleration due to gravity.
h=v_0 ^2sin^2θ/2g
The value of g is 9.8m/s^2. We want to find the height of the ball for different values of θ if v_0 is 47 meters per second. Let's start with substituting these values into the formula and simplifying.
The height of the ball for θ=30^(∘) is approximately 28.2 meters. Let's now find the heights for other angles.
θ
2209sin^2θ/19.6
h=2209sin^2θ/19.6
Approximation
45^(∘)
2209sin^2 45^(∘)/19.6
56.352040...
56.4
60^(∘)
2209sin^2 60^(∘)/19.6
84.528061...
85.5
90^(∘)
2209sin^2 90^(∘)/19.6
112.704081...
112.7
b
We will now graph the equation on our graphing calculator. We want to graph it when g= 9.8 and v_0= 47. We will use the simplified form we have obtained before.
h=( 47) ^2sin^2θ/2( 9.8)
⇕
h=2209sin^2θ/19.6
For the purpose of graphing we will change h for y and θ for x.
y=2209sin^2x/19.6
In this case we will consider the x-values between 0 and 180. Also in the previous part of exercise, we have found that for x=90, y≈ 112.7. Therefore, the maximum value for y in the window needs to be significantly bigger, let's say 150. The minimum value for y will be 0. We will now push WINDOW and resize it.
Now, let's graph the equation. Press the Y= button and type the equation in the first row. Having written the equation, push GRAPH to draw them.
c
Now, we want to prove that the formula is equivalent to the given one.
v_0 ^2tan^2θ/2gsec^2θRecall the definition of secant and one of the Quotient Identities considering tangent.
secθ = & 1/cosθ
tanθ = & sinθ/cosθ
