Question

In 1939 or 1940, Emanuel Zacchini took his human-cannonball act to an extreme. After being shot from a cannon, he soared over three Ferris wheels and into a net. Assume that he is launched with a speed of 26.5 m/s and at an angle of 53.0° and neglect air drag.

a) Treating him as a particle, calculate his clearance over the first wheel.
b) If he reach maximum height over the middle wheel, by how much did he clear it?
c) How far from the cannon should the net's center have been positioned?

Answer 1

See attached for accompanying picture.

Answer 2

sooo what's the question?

since the guy is being launched at an angle, break that up into x and y components
and break up the initial velocity into x and y components as well

Answer 3

x coordinate motion has no acceleration thus its
\[x=v_{ox}t\]
y coordinate motion has an acceleration thus you need to use
\[y=-\frac{1}{2}at^2+v_{oy}t+x_{oy}\]

Answer 4

@completeidiot I retyped the question to include ... the question. Lol...
I've tried doing it but my answers come up different from the real answers, apparently:
a) 5.3m b) 20m c) 69m

Answer 5

gimme a sec

Answer 6

Isn't b associated with the equation Hmax = \[\frac{ v _{0^{2}}\sin2\theta }{ 2g } \]

Answer 7

I tried plugging in but my answer is always off

Answer 8

dunno, i always just solve for the time it reaches max height using derivatives, then plugging that back into the height equation, then subtracting the height of wheel

Answer 9

lemme work it out

Answer 10

I know derivatives but we're just working with trig here so if I theoretically didn't know calc...

Answer 11

meh fine -b/2a = vertex

Answer 12

well thats the t coordiate so plug that back into heigh equation to find actual height, then subtract height of wheel

Answer 13

v_y=26.5 sin(53)=21.163
v_x=26.5 cos(53)=15.948
screw significant figures

x_o =0
because you're landing on a net also 3 m tall

y=-.5(9.8)t^2+ 21.163*t

y=0=-4.9t+21.163
t=4.31

vx*4.31= total distance travelled=68.88m

clearing the first whell

15.948m=vx
determine time when he reaches 23 meters horizontally
23/15.948=1.44218 sec
plug that in for t and determine height

h(1.44218)=20.329

determine vertex
cheating by using a graphing calculator
t=2.1594s h=22.85m

using the time value of the vertex, determine horizontal distance travelled

34.438m away from cannon

Answer 14

oh wait
you dont need to know how far away the middle wheel was

Answer 15

... slightly confused but thank you

Answer 16

also did you take into account that the initial vertical displacement was 3m from ground level @completeidiot

Answer 17

you have a vertical displacement of 3, but the net is also a vertical displacement of 3

3-3 = 0

Answer 18

you're landing on the net, not hte ground

Answer 19

oh, right

Answer 20

it would be quite painful to hit the ground at about 26.5m/s

Answer 21

actually it would be faster, but only by a little

Answer 22

lol

Answer 23

let me try and absorb this... (I got put into AP Physics C when I wanted Physics B, lol) a medal for your efforts

Answer 24

if you dont understand anything, feel free to ask
actually send me a message instead

Answer 25

for b - I was given an angle so I guess I'm expected to use it... I would think it would be [(26.5)^2*sin2(53)]/2g but then I get a height of 34.4 which, minus 18, is 16.4m and not 20m

Answer 26

whatever, I'll just put that down. I'm pretty tired