The graph below shows the height of a projectile t seconds after it is launched. If acceleration due to gravity is –16 ft/s2, wh
ich equation models the height of the projectile correctly?
1 answer:
Answer:
see attached
Step-by-step explanation:
The x-coordinate of the vertex is given by ...
x = -b/(2a)
for f(x) = ax^2 +bx +c.
We see that the vertex is at x = 1, and we are given a = -16. Then we can find "b" from ...
1 = -b/(2(-16))
32 = b
We know the y-intercept of the equation is 5 (the starting height), so the equation must be ...
h(t) = -16t^2 +32t +5 . . . . . matches the last choice
You might be interested in
Assuming you want the equation in slope intercept form, you first need to use slope formula to find the slope.
Plug in the coordinates that you have, (3,-1) and (4,7), for the x and y values respectively.
Which reduces to 8/1=8.
The slope of your equation is 8.
Slope intercept form is: y=mx+b. m is your slope and b is your y-intercept, x and y stay unchanged.
Plug the slope in:
y=8x+b.
Now, you can plug one of the sets of coordinates into the x and y of the equation to find b. I'm using (3,-1).
-1=8(3)+b
-1=24+b
Subtract 24 from both sides.
-25=b
Now plug this into the equation!
y=8x-25 is your final equation.
I hope this helps :)