A new circus act features a large frictionless cannon with an 2500 N/m spring mo

Question

A new circus act features a large frictionless cannon with an 2500 N/m spring mounted at the bottom of a long barrel which makes an angle of 37.0o with the horizontal. A 100 kg acrobat crawls inside the barrel with his feet against the cannon's spring which has been compressed 4.00 m. When released, the acrobat travels 10.0 m up the inclined barrel.
a) Find the speed of the acrobat when he reaches the top of the barrel. Use energy methods.
b) Upon reaching and exiting the top of the barrel, the clown heads for a tub of water located horizontally from and 12.0 m vertically below the canon

Explanation / Answer

(a) When the acrobat travels 10m up the barrel,

then height covered = 10 sin37 = 6 m

Conserving energy,

0.5 kx^2 = mgh + 0.5 mv2cannon

=> (0.5 x 2500 x 4^2) = (100 x 9.8 x 6) + (0.5 x 100 x v^2)

=> vcannon = 16.8 m/s

(b) Using energy conservation,

0.5 m v2cannon = 0.5 m v2tub - mgh

vtub = sqrt[v2cannon + 2gh]

= sqrt[16.8^2 + (2 x 9.8 x 12)]

= 22.75 m/s

(c) For time of flight, the vertical displacemet from the cannon is y = -12 m.

Initial vertical velocity = vcannon sin37 = 10.08 m/s

y = (vcannon sin37)t - 0.5gt^2

-12 = 10.08t - 4.9t^2

4.9t^2 - 10.08t -12 = 0

t = 2.9 seconds

Horizontal distance covered = horizontal velocity x time of flight

= vcannon sin37 x t

= 16.8 x 0.8 x 2.9

= 38.976 m

= 40 m (approx)

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