what is the equation of the flight and the max ht attained?

Question

. (Human cannonball) In the 1940s, the human cannonball stunt 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 horizontal distance traveled was 175 feet. When the cannon is aimed at an angle of 45 degrees, an equation of the parabolic flight (see the picture) has the form y = ax^2 + x + c

Answer 1

Let the tip of the cannon be at x = 0. At that point, y = 15; then c = 15. The distance travelled is the (positive) value of x when y = 0. Let that be s; then

as^2 + s + 15 = 0.

You are given that s = 175; then

a*175^2 + 175 + 15 = 0

a = - 190/(175^2) = -.00624

The equation is then y = - 0.00624x^2 + x + 15

The max height occurs when dy/dx = 0:

-2*.0064*x +1 = 0

x = 1/ (2*0.0064) = 78.125 ft

Put this into the equation for y to get max ht:

-0.0064*78.125^2 + 78.125 + 15 = 54.06 ft

Answer 2

to find the equation that represent the path of the man, you need first to find the initial speed of the man.


let V be the intial speed. Because the man is fired at an angle, there are 2 speed components. One is in the x-direction and the other one is in the y-direction. The man accelerates vertcial and travels at a constant speed horizonally.

Break down the initial speed
Vx = Vcos45
Vy = Vsin45

Break down the intial speed
Vx = Vcos45
Vy = Vsin45

find the time it takes the ball to hit the ground 175m away horizonally

x = vt
175 = Vcos45 * t
t = 175 / Vcos45

Xf = .5at^2 + Vt + Xi

Xf = final position (0m)
Xi = initial position (15ft)
a = acceleration
t = time
V = speed in y-derection

we just found the time, which is 175 / Vcos45, plug it in

0 = .5(-32)(175 / Vcos45)^2 + Vsin45 * (175 / Vcos45) + 15

-15 = -16(175/ Vcos45)^2 + 175

-190 = -16(175 / Vcos45)^2

11.875 = (175 / Vcos45)^2

3.446 = 175 / Vcos45

3.446 * Vcos45 = 175

V = 175 / (3.446cos45)

V = 71.82m/s


Now that we found the intial speed, use it to find the equation that represent the path of the man

Vx = 71.82cos(45) = 50.78
Vy = 71.82cos(45) = 50.78


x = vt
x = (50.78)t
t = x / 50.78

y = .5at^2 + vt + yi
y = .5(-32)(x / 50.78)^2 + (50.78)(x / 50.78) + 15

y = -0.006204x^2 + x + 15

The equation above represent the path of the man. The maximun height is the y valve of the vertex

Answer 3

Sorry, no picture.
Did you mean ax^2 +bx +c instead of ax^2 +x +c?

The max height will occur when x=-b/2a.

You really have not supplied sufficient data.
The trajectory is given by:
y = -gx^2/(2Vo^2cos^2z) + xtanz, where;
g is the gravitational constant,
Vo is the initial velocity,
z is the inclination angle,
y is the height at anyy point x,
x is the horizontal distance traveled

Parametrically, the equations are
x = (Vo cos z) t and y = -.5gt^2 + (Vosin z)t, where t = time in seconds.

Answer 4

what are you trying to find?

Answer 5

so? whats your point?