Question

Please help me. I'm really confused and could use some help with precalc.

Answer 1

Using formula h=(v/16)sinxcosx; when the velocity is 60 ft per second, what is the maximum horizontal distance (h) possible and at what angle (x) does this occur?

Answer 2

sin(2x) = 2sin(x)cos(x)
sin(x)cos(x) = sin(2x) / 2

h = (v/16) * sin(x)cos(x) = (v/16) * sin(2x) / 2
h = v/32 * sin(2x)
v = 60 ft / sec
h = 60/32 * sin(2x).

To maximize h, sin(2x) has to be maximum.

What is the maximum value of a sine function?
And at what angle does this maximum occur?

Answer 3

h = 60/32 * sin(2x)
The maximum value of a sine function is 1.
Therefore, the maximum value of h is: 60/32 * 1 = 60/32 = 15/8 feet = 1 7/8 feet.

The sine function attains the maximum value of 1 when the angle is 90 degrees.
Therefore, sin(2x) = 1 when 2x = 90 degrees
or x = 45 degrees.

Answer 4

But can you check if what you posted in the first reply is correct?
h = (v/16)sinxcosx
I think it should be:
h = (v^2/16)sinxcosx (v squared instead of v).

Answer 5

IDK why I keep posting replies when it does not look like you will reply.