Answer: a) BC = 1386.8 ft
b) CD = 565.8 ft
Step-by-step explanation:
Looking at the triangle,
AD = BD + 7600
BD = AD - 7600
Considering triangle BCD, we would apply the the tangent trigonometric ratio.
Tan θ = opposite side/adjacent side. Therefore,
Tan 24 = CD/BD = CD/(AD - 700)
0.445 = CD/(AD - 700)
CD = 0.445(AD - 700)
CD = 0.445AD - 311.5 - - - - - - - -1
Considering triangle ADC,
Tan 16 = CD/AD
CD = ADtan16 = 0.287AD
Substituting CD = 0.287AD into equation 1, it becomes
CD = 0.445AD - 311.5
0.287AD = 0.445AD - 311.5
0.445AD - 0.287AD = 311.5
0.158AD = 311.5
AD = 311.5/0.158
AD = 1971.52
CD = 0.287AD = 0.287 × 1971.52
CD = 565.8 ft
To determine BC, we would apply the Sine trigonometric ratio which is expressed as
Sin θ = opposite side/hypotenuse
Sin 24 = CD/BC
BC = CD/Sin24 = 565.8/0.408
BC = 1386.8 ft
81.6. Just entered it in a caculator.
Answer:
C
Step-by-step explanation:
All of these equations will be set up as: h(t) =
+v₀t + h₀ where g represents gravity, v₀ represents initial velocity, and h₀ represents initial height. When working with ft/sec, g = 32. So, -g/2 = -16
1a) Length of time to reach its maximum height means you are looking for the x-value of the vertex (aka Axis Of Symmetry).
h(t) = -16t² + 160t
AOS: x =
=
= 5
Answer: 5 sec
1b) Length of time to fall to the ground means you are looking for the x-intercept when height (y-value) = 0.
h(t) = -16t² + 160t
0 = -16t² + 160t
0 = -16t(t - 10)
0 = -16t 0 = t - 10
t = 0 t = 10
t = 0 is when it started, t = 10 is when fell to the ground.
Answer: 10 sec
2c) Same concept as 1a
h(t) = -16t² + 288t
AOS: x =
=
= 9
Answer: 9 sec
2d) Same concept as 1b
h(t) = -16t² + 288t
0 = -16t² + 288t
0 = -16t(t - 18)
0 = -16t 0 = t - 18
t = 0 t = 18
Answer: 18 sec
3e) Same concept as 1a
h(t) = -16t² + 352t
AOS: x =
=
= 11
Answer: 11 sec
3f) Same concept as 1b
h(t) = -16t² + 352t
0 = -16t² + 352t
0 = -16t(t - 22)
0 = -16t 0 = t - 22
t = 0 t = 22
Answer: 22 sec