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posted by  Morty23 on 7/24/2008 1:22:50 PM  |  status: Live  

Newtons Laws of Motion

Course Textbook Chapter Problem
Algebra Based Physics Fundamentals of Physics Extended (8th) by Resnick, Halliday, Walker 5 49
Question Details:
The Zacchini family was renowned for their humancannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of 69m. He was propelled inside the barrel for 5.2m and launched at an angle of 53 degrees. If his mass was 85kg and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at 53 degrees. Neglect air drag. )
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posted by Jamil on 7/24/2008 2:57:35 PM  |  status: Live
Asker's Rating: Helpful   
Morty23's comment:
"found the acceleration but not the force. did not answer the question."
Response Details:
Let, v= velocity at which the person came out of the cannon.
horizontal velocity = [v * (cos 53) ]m/s = 0.602v
vertical velocity = [v * (sin 53) ]m/s = 0.799v
Let, t = time for which the person was in the air.
∴0.602 * v * t = 69    [speed * time = distance]
or, v = 69/(0.602 t).......................................................  [equation 1]
s = u t + 0.5 a t2
or, 0 = 0.799v t + 0.5 (-9.81) t2
or, 0 = [(0.799) * (69 / 0.602t) * t ]- 4.905 t2
or, 0 =  91.58 - 4.905 t2
or, t2 = 18.67
or, t = 4.32 s
From equation 1:
v = 69 / (0.602 * 4.32) = 26.532 m/s
v2 = u2 + 2 a s
or, 26.5322 = 0 + 2 a * 5.2
or, a = 67.69 m/s2
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posted by PDavis88 on 7/24/2008 3:06:18 PM  |  status: Live
Asker's Rating: Helpful   
Morty23's comment:
"The solution seems correct but the answer in the back of book is 6.4*10^3 N. thank for getting me on the right track."
Response Details:
Basically what you are trying to do is first find the velocity at which he leaves the cannon and then use that to find the acceleration.
Start by making a diagram:


V=velocity at exit
Vx=velocity at exit in x direction
Vy=velocity at exit in y direction

Now you can make 2 equations for the motion:
Vx=Vcos53
Vy=Vsin53

You know the x-distance, so you can use kinematics to form an equation for it:
69=Vcos(53)*t

Now make an equation for the velocity in the y direction
0=Vsin53+a*(t/2)
t/2 because it is half of the total time in the air
a=-9.81 for gravity

Solve both equations for t

set them equal to each other:

Now solve for V
V=26.54 m/s

Now you can use a kinematic equation to find the acceleration until it hit that point:
vf2=vi2+2ad
26.542=0+2*a*5.2
solve for a
a=67.73 m/s2

Finally, just use the equation F=ma to find the force
F=85*67.73
F=5757N
Thank you for rating me!
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