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Rocket Fight Profile

Physics
by

Francie Bernstein

on 15 January 2013

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Transcript of Rocket Fight Profile

My Rocket Flight
Profile By: Francie Bernstein
Period-4 Newton's Laws Pre-Launch Lift Off Max Altitude Descent Cruise During this unit and project, I learned all about Newton's laws of motion and they applied to the launch of the rocket. The first law states that an object at rest stays at rest unless it is acted upon by an unequal force. His second law states that in a presence of a net force, an object experiences an acceleration. Newton's third law states that any two objects attract each other with the universal force of gravity. In the pre launch stage, the rocket experiences Newton's first law and third law of motion. The rocket stays at rest, but the lauch pad has an equal, but opposite force acting on it. The normal force supports the object, causing it to stay at rest. When the rocket is moving up, the force of gravity slows it down According to Newton's first law, if there was no gravity, the rocket would continue to go on for ever and ever. But because gravity is in effect, it's bringing the rocket down V = 90.2m/s
V = 0m/s
t= 0.5sec
m= 0.017kg Touchdown! During this stage, the rocket fell at a constant speed and the air resistance is equal to the force of gravity. g F = (-9.8 m/s )(0.017 kg)= 0.17 N 2 In the lift off stage, the rocket experiences Newton's 2nd and 3rd law V = 0
mass = 0.017 kg
Impulse = 2Ns
t = 0.7 sec
a = ? 0 V = ?
h = ? f (see more calculations in circle #5) Lift Off (continued) I = Ft = m (V -V )
2Ns= (0.017 kg)(V - 0)
2Ns/ 0.017 kg = V
117.6 m/s = V f 0 f f f a = V -V / t
= 117.6m/s - 0 /0.7 sec
= 167 m/s f 0 2 a = a + a
= 167 m/s + (-9.8m/s ) net E g 2 2 * these calculations do not include the force of gravity https://sites.google.com/a/lvcp.org/physicsland/projects/rocket-videos/francie.MP4?attredirects=0&d=1 Lift Off (continued) F = 0.17N N F = mg
= (0.017 kg) ( -9.8 m/s )
= 0.16 N g 2 F = ma
= (0.017 kg) (167 m/s )
= 2.84 N E E 2 F = F + F
= 2.84N + (-0.16N)
= 2.68N net E g Lift Off (continued) F = ma
F / m = a
268N / 0.017 kg = 157 m/s net net net net 2 t*a = V -V *t / t
= 157m/s (0.7sec)
= 110m/s f 0 * = multiplication 2 x= 1/2 ( V + V ) t
= 1/2 ( 110m/s) (0.7sec)
= 38m f 0 a + air resistance = -11.7m/s g 2 m= 0.017kg
V = 110m/s
V = 0 m/s
a= -9.8 m/s
h= ?
t = ? 2 f 0 a = V - V / t
t = V - V /a
t= 0 - 110 m/s / -9.8m/s
t= 11 sec f 0 f 0 x= x + v t + 1/2 at
x= 38m + (110m/s) (11sec) + 1/2 (-9.8m/s ) (11sec) V = 0m/s
a = -9.8m/s 2 h + h
38m + 541m = 579m burn cruise V = d/t
= 541m /6sec
= 90.2 m/s * It took approximately 6 sec to hit the ground f 0 F = m (V -V /t)
= 0.017kg (0-90.2m/s / 0.5sec)
= -3.1N * Because the rocket is applying a downward force, there is a negative answer
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