Recall that the equation for horizontal distance "h" in feet of a projectile with initial velocity v0 and initial angle theta is given by h=(v0^2/16)sin theta cos theta.
a.) Assume the initial velocity is 60 (feet/second). What initial angle will you need to ensure that the horizontal distance will be exactly 100 feet?
b.) Assume the initial velocity is 60 feet/second. What is the maximum horizontal distance possible and at what angle does this occur?

Question

Recall that the equation for horizontal distance "h" in feet of a projectile with initial velocity v0 and initial angle theta is given by h=(v0^2/16)sin theta cos theta.  
a.) Assume the initial velocity is 60 (feet/second). What initial angle will you need to ensure that the horizontal distance will be exactly 100 feet? 
b.) Assume the initial velocity is 60 feet/second. What is the maximum horizontal distance possible and at what angle does this occur?

amistre64 · Answer

that doesnt look like a horizontal distance formula to me ...  at least not a standard one

amistre64 · Answer

the horizontal distance (along the x axis) of a projectile tends to be:  x = vo t cos(a)

i spose they are solving the vertical to sub into t maybe?

y = -gt^2 +vo t sin(a) + yo

0 = -gt^2 +vo t sin(a) + yo

t = vo sin(a) +- sqrt(vo^2 sin(a)^2 + 4g(yo)) ] /2g

x = vo  cos(a) [vo sin(a) +- sqrt(vo^2 sin(a)^2 + 4g(yo)) ] /2g

x = vo^2  sin(a)cos(a)/ 2g +- vo cos(a) sqrt(vo^2 sin(a)^2 + 4g(yo)) ] /2g

ok, if we let the sqrt part = 0 ... we can determine what they have provided, but  g = 32/2

x = vo^2  sin(a)cos(a)/32

amistre64 · Answer

http://en.wikipedia.org/wiki/Range_of_a_projectile yeah, its spose to be /32 not /16

anonymous · Answer

Well, the equation set up actually turns out as 100= (60^2/16)cosxsinx, and I'm pretty sure I'm solving for x... as in what to input into sin and cos. Though the 32 thing is weird... The problem was written as vo^2/16...

amistre64 · Answer

i see a mistake i made

2 sin(a) cos(a) = sin(2a)

soo   the /16 is right for the sin cos version

sin(2a)/32 is better for what we have to accomplish tho

amistre64 · Answer

$$h=\frac{(v_o)^2}{32/2}sin(a)cos(a)=\frac{(v_o)^2}{32}sin(2a)$$
$$\frac{32h}{(v_o)^2}=sin(2a)$$
$$sin^{-1}(\frac{32h}{(v_o)^2})=2a$$
$$\frac12sin^{-1}(\frac{32h}{(v_o)^2})=a$$

amistre64 · Answer

http://www.wolframalpha.com/input/?i=arcsin%2832%28100%29%2F%2860%29%5E2%29%2F2 this gives us radians, which you then need to convert to degrees

anonymous · Answer

... 
Hokay. Think that did it. Hopefully it worked out. I'm getting 100=(3600/32)(sin(2(31.367)))

anonymous · Answer

Yep

anonymous · Answer

There we are

anonymous · Answer

Thank you