Question

Q:The horizontal distance traveled by a projectile fire with initial angle theta and initial velocity 60 m/s can be modeled by the equation

h=(7200/9.8) sin theta cos theta

Choose the angle where the horizontal distance is at a maximum?"
A:
pi/6
pi/4
pi/2
pi/3"

So is the equation have to be 60x(7200/9.8)sin theta cos theta? or graph it?

Answer 1

can u help me here?

what is the center of the circle that has a diameter whose endpoints are (2,7) and (-6, -1)?

ans: (4,3), (-2, 4), or (-2,3)?

Answer 2

4,4,3 I believe

Answer 3

because its just looking for mid point of the diameter

Answer 4

for your question or mine?

Answer 5

your's its -2,3

Answer 6

the center is

(2-6) / 2 , (7-1)/ 2

Answer 7

ladrybro is correct

Answer 8

thanks!

Answer 9

find the distance between the points (-4,-5) and (3,-1)

Answer 10

the anggle givingthe maximum horizontal distance in ladybro's question is
the value of theta where sin theta cos theta is a maximum

Answer 11

you can use calculus to find this value which is pi/4 ( 45 degrees)

Answer 12

okay so input it into the theta the 45 degree?

Answer 13

to find the value of the max distance plug 45 degrees into your formula

Answer 14

I got 367.3469.. with the equation above

Answer 15

h=(7200/9.8) sin theta cos theta
is the correct formula

not 60 * h=(7200/9.8) sin theta cos theta

Answer 16

h = (7200/9.8) * (1/sqrt2) * (1/sqrt2)
= 3600 / 9.8

Answer 17

your answer is correct