Water Rocket Lab Analysis

Rocket Project Calculation/Questions/Analysis 1. Calculate the sign focal ratio at which it left the launch fatten out The ground aim was zero and the flower obtained was 100 m (assuming the missile travels perpendicularly to the sur represent of the ground). The stop number is zero. We have snatch of five variables require to solve for initial hurrying. V= 0 m/s at top of climax, the rocket has no velocity as it turns to face the ground x = 100m the so-called height of the rocket from climax to ground level. This was calculated utilize trigonometry: the angle out from launch pad and exceed away from launch pad. a = -10 m/s^2 constant acceleration of gravity sign velocity is form utilise the following equation. V2=Vo2+ 2ax V is lowest velocity, Vo is initial velocity, a is acceleration, and x is distance. 0=Vo2+ 2-10100 Vo2=2000 Vo2=44.72 m/s initial velocity is 44.72 meters per second. 2. Time the rocket, from launch to apogee, and confi rm the height of the rocket at apogee (you may NOT use your consequence for image one). There be enough factors to confirm the height of at apogee without using the initial velocity from worry one. t = 3.88 s this was recorded from the moment the rocket was launched to apogee using a stopwatch.

V= 0 m/s at top of apogee, the rocket has no velocity as it turns to face the ground a = 10 m/s^2 constant acceleration of gravity To calculate the height, initial velocity is placed using the following equation. V=Vo+ at V is velocity, Vo is initial velocity, a is acceleration, and t is term. 0=Vo+ -103.88 Vo=38.8 m/s To find the height at apogee: ! ?x=Vot+ 12at2 x is distance from launch to apogee. ?x=3.88(38.8)+ 12(10)(3.88)2 ?x=75.27 m whence the height given in paradox 1 and the time calculated in problem 2 does not check into and does not confirm. 3. Draw a diagram of the rocket- label the move and give a short description for why they are there. The rocket is made up of three components the top...If you want to drop dead a full essay, order it on our website: BestEssayCheap.com

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