A)use the information to find the equation of the flight.?

Question

in the 1940's, the human cannonball stut was performed regularly by Emmanuel Zacchini for the Ringling Brothers and Barnum & Bailey Circus. The tip of the Cannon rose 15 feet off the ground and the total horizontel distance traveled was 175 feet. When the cannon is aimed at an angle of 45 degrees, an equation of the parsbolic flight has the form y=ax(squared)+x+c

Answer 1

You can solve this problem with a bit of physics and a few observations. Yes, the flight path will have a parabolic trajectory. Notice a few things:

1. At x=0 (no distance traveled), the stuntman's position is 15 ft above the ground. Thus:

y(0) = a(0)² + (0) + c =15 ft

Therefore, c = 15 ft.

2. At x = 175 ft (full-range), the stuntman's height is 0 ft above the ground because he's landed. Thus:

y(175) = a(175)² + 175 + 15 = 0

We can now solve for the value of a as:

a(175)² + 175 + 15 = 0

a(175)² = -175 - 15

a = -(175 + 15) / (175)²

a = - .0062

Therefore, your equation of flight is:

y = (-.0062)x² + x + 15


Hope that makes sense :)

Answer 2

For this whole equation assuming stuntman is being shot left to right. [ --->]
Given when y = 15 (height of cannon barrel end), x= 0 ft..
[next is assuming he landed on the ground]
Given when y = 0 (ground level), x = 175 ft.
C = 15, height of cannon.
y = ax² + x + C
y = ax² + x + 15 [plug in x=175, y=0 to find a]
0 = a(175)² + 175 + 15
0 = 30625a + 195
a = -39/ 6125
y = (-39/6125)x² + x + 15 <- this is your parabolic equation.
EXTRAS
y' = (-78/6125)x + 1
0 = (-78/6125)x + 1
-1 = (-78/6125)x
x = (6125/78)
x = 78.53 ft
So when he was at a distance of 78.53 ft from the barrel, he was at his highest point. Meaning at a distance of 157.05 ft, he cross the plane were y = 15.

Answer 3

The equation of the path of a projectile launched from a height of 15 units at an angle of 45 deg is
y = -g*(x^2)/(V^2) + x + 15, where g is the gravitational constant which, if I remember correctly, is 32 in obsolete units. (i.e. feet etc)

The information is that y = 0 when x = 175, so sub all this in and solve for V, put that value of V back into the equation and you have the required equation.

Answer 4

Assume the cannon was at 45 degrees, then
initial speed in horiz. and vert. dir. are equal, or
ux=uy=u
S=total distance travelled in time t,
then u=S/t
After time t, the y distance is -15 ft, with
g=32.2 ft/sec/sec, acceleration due to gravaity, then using
-15=ut+at^2/2=ut-gt^2/2
Substitute S=175, t=(S/u), g=32.2,
u(S/u)-32.2(S/u)^2/2=-15
175-16.1(175/u)^2=-15
175/u=sqrt((175+15)/16.1)
u=175/3.435=50.942
substituting x=ut, or t=x/u,
y=ut-gt^2/2=x-(g/2u^2)*x^2
Therefore if y=0 and x=0 at t=0,
a=-(32.2/(2*50.942^2)
=-0.00620
=-1/161.19
c=0
(please check numerical calcultions)