The range of a projectile that is launched with an initial velocity v at an angle of with the horizontal given by R=v^2/g
sin 2 theta
where g is the acceleration due to gravity or 9.8 meters per second squared. If a projectile is launched with an initial
velocity of 13 meters per second, what angle is required to achieve a range of 15 meters?

the answer is 844

step-by-step explanation:

it is 844 because if you add 836.50 + 7.5 you will get 844

answer: h=5.6 i don't really know how to put into words, so i though i would let you do that. i hope i was able to some

step-by-step explanation:

18 = 6(h)+1/2

18 = 6h+1/2

18-1 = 6h/2

17 = 3h

17/3 = h

h = 5.6

b. 18

step-by-step explanation:

so the equation for this is:

ab^2=ac⋅ad

ab being the top length of 6 and ac being the length of 2 and ad being the length of x. let's plug into the equation:

6^2 = 2x

now lets just solve for x:

36 = 2x

18 = x

i hope that !