Your initial velocity component up which you are calling Vo is 1750 sin 10
so
Vo = 304 ft/s
( I can hardly believe you are using feet not meters but well here g will be 32 ft/s^2)
Your distance down range, x = u t
where your constant horizontal speed u = 1750 cos 10 (that is part b by the way)
so
u = 1723 ft/s
In your equation if you fire at ground level the Hi on the right is zero. That is the initial height
so
h = -16 t^2 + 304 t
now if we used physics and calculus we would zip ahead, but I suspect you are doing algebra 2 so we will do completing the square on the parabola that will result from plugging in x = u t so t = x/u
h = -16 (x^2/u^2) + 304 (x/u)
since u = 1723
1723^2 h = -16 x^2 +304(1723) x
complete the square on that to find the vertex of the parabola. That is max h and x at max h
Of course t at max h = x/u
Then twice that x and twice that time is when h = zero, ground
that gives you the range and time in the air.
I am going to leave it to you to discover that max range is when you fire at 45 degrees from horizontal.
By the way a physicist would use:
v = Vo - 32 t
and when v = 0, you are at the top
We fire a missile from ground level at 1750 ft/sec. at launch angle of 10 degrees. h(t) =-16t^2 + vot +h
a. find a function for height as a function
b. find a function for distance from launch point as a function of time.
c. when will the missile reach max height?
d. What is max height?
e. how long will the missile be in the air?
f. how far will it travel?
g. find a function for distance travelled as a function of launch angle
h what is max distance? what launch angle maximizes distance?
find a formula for launch angle as a function of distance
1 answer
1 answer