Water Rocket Lab Analysis

Rocket Project Calculation/Questions/Analysis 1. Calculate the sign focal ratio at which it left the launch roll The ground aim was zero and the flower obtained was 100 m (assuming the missile travels perpendicularly to the sur compositors case of the ground). The upper is zero. We have strangler of five variables require to solve for initial speeding. V= 0 m/s at top of climax, the rocket has no velocity as it turns to face the ground x = 100m the alleged(a) height of the rocket from climax to ground level. This was calculated exploitation trigonometry: the angle out from launch pad and infinite away from launch pad. a = -10 m/s^2 constant acceleration of gravity sign velocity is undercoat development 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 ap ogee, and confirm the height of the rocket at apogee (you may NOT use your consequence for fragmentise one). There argon 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 date. 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 indeed the height given in paradox 1 and the time calculated in problem 2 does not yoke and does not confirm. 3. Draw a diagram of the rocket- label the part and give a short description for why they are there. The rocket is made up of three components the top...If you want to study a full essay, order it on our website: BestEssayCheap.com

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