I think it is always true. I am not sure but I hope I'm right
Answer:
I got B or number 2
Step-by-step explanation:
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>
Plug 4 for x and solve.
y = -3/4 * 4 + 3
y = -12/4 + 3
y = -3 + 3
y = 0
So the missing coordinate is 0.
Answer:
Or if you want with the value of h too.
Step-by-step explanation:
Find the value of h and k by using the formula.
From y = x²-2
Substitute these values in the formula.
Therefore, h = 0.
Therefore, k = - 2.
From the vertex form, the vertex is at (h, k) = (0,-2). Substitute h = 0, a = 1 and k = -2 in the equation.
These type of equation where b = 0 can also be both standard and vertex form.