The question involves the concept & equations associated with projectile motion.
Given:
y₁ = 1130 ft
v₁ = +46 ft/s (note positive sign indicates upwards direction)
t = 6.0 s
g = acceleration due to gravity (assumed constant for simplicity) = -32.2 ft/s²
Of the possible equations of motion, the one we'll find useful is:
y₂ = y₁ + v₁t + 1/2gt²
We can just plug and chug to define the equation of motion:
<u><em>y = (1130 ft) + (46 ft/s)t + 1/2(-32.2 ft/s²)t²</em></u>
<em>(note: if you were to calculate y using t = 6.0 s, you'd find that y = 826.4 ft, instead of 830 ft exactly because of some rounding of g and/or the initial velocity)</em>
Answer:
B
Step-by-step explanation:
2.3 is being cancelled out.
Let r be a radius of a given circle and α be an angle, that corresponds to a sector.
The circle area is
![A=\pi r^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2)
and denote the sector area as
![A_1](https://tex.z-dn.net/?f=A_1)
.
Then
![\dfrac{A_1}{A}= \dfrac{\alpha}{2\pi}](https://tex.z-dn.net/?f=%20%5Cdfrac%7BA_1%7D%7BA%7D%3D%20%5Cdfrac%7B%5Calpha%7D%7B2%5Cpi%7D%20)
(the ratio between area is the same as the ratio between coresponding angles).
![A_1=\dfrac{\alpha}{2\pi} \cdot A=\dfrac{\alpha}{2\pi} \cdot \pi r^2= \dfrac{r^2\alpha}{2}](https://tex.z-dn.net/?f=A_1%3D%5Cdfrac%7B%5Calpha%7D%7B2%5Cpi%7D%20%5Ccdot%20A%3D%5Cdfrac%7B%5Calpha%7D%7B2%5Cpi%7D%20%5Ccdot%20%5Cpi%20r%5E2%3D%20%5Cdfrac%7Br%5E2%5Calpha%7D%7B2%7D%20)
.
The second one should be the answer