EXAMPLE 6 A projectile is fired with muzzle speed 170 m/s and an angle of elevation 45° from a position 20 m above ground level. Where does the projectile hit the ground and with what speed? SOLUTION If we place the origin at ground level, then the initial position of the projectile is (0, 20) and so we need to adjust the parametric equations of the trajectory by adding 20 to the expression for y. With v0 = 170 m/s, α = 45°, and g = 9.8 m/s2, we have

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

where is the graph

step-by-step explanation: