A water rocket, launched from the ground, rises verticallywith acceleration of 30 /s2 for 1 s when it runs out of fuel. Disregarding air resistance, how high will the rocket rise? Comments

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Jaylen Fountain · Accepted Answer

1. The distance traveled during the first second (positiveacceleration): △y=v0t+12at2=0+12(30)(12)=15m The speed after this first second: △v2=2a△y v2=v02+2a△y=0+2(30)(15) v=30 m/s The additional distance traveled (acceleration of g in thenegative direction). At the peak of the flight, the velocityis zero: △v2=2a△y △y=△v22a=0−v22(−g)=v22g=3022(9.8)=45.92 Other equation(s) of motion could have also been used to findthis distance. Total distance traveled: y=15+45.92=60.92m 2. d=v0t+12at2 the initial velocity is zero. I am going to assume that theacceleration is the average acceleration of the rocket, that is thevertical acceleration minus the acceleration of gravity. I am also assumeing that the acceleration is in meters/secondsquared. this means that: d=0+12(30)(1)=15m

Jeffrey Jordon · Accepted Answer

v=at=30msd=v2−vo22a=−(302)(2x−9.8)d=45.92m+h (h is the initial height before it ran out offuel)  h=0.5(30)(12)=15 md=45.92+15=60.92m

nick1337 · Accepted Answer

Using the equation v=v0+at, we can find the final velocity of the rocket when it runs out of fuel. Since the rocket starts from rest (v0=0), the equation simplifies to v=at. Substituting the given values, we have v=(30ms2)·(1s)=30 m/s.Next, we can use the equation s=v0t+12at2 to find the displacement of the rocket during the 1 second interval. Again, since the rocket starts from rest, the equation simplifies to s=12at2. Substituting the given values, we have s=12(30ms2)·(1s)2=12·30&amp;nbsp;m=15 m.Therefore, the rocket will rise to a height of 15 meters when it runs out of fuel, disregarding air resistance.

Vasquez · Accepted Answer

Step 1: Given:Δy=vit+12at2where Δy is the change in height, vi is the initial velocity, a is the acceleration, and t is the time.Step 2: Given that the rocket rises vertically with an acceleration of 30m/s2 for 1s, we can substitute the values into the equation:Δy=0·1+12·30·(1)2Simplifying the equation, we get:Δy=12·30mTherefore, the rocket will rise to a height of 15m.

RizerMix · Accepted Answer

Answer: 15 metersExplanation:s=ut+12at2where:- s is the displacement (height in this case),- u is the initial velocity (0 m/s since the rocket starts from rest),- a is the acceleration (30 m/s2),- and t is the time (1 s).We want to find the height the rocket will reach, so we need to calculate s using the given values. Plugging in the values into the equation, we have:s=0·1+12·30·(1)2Simplifying the equation gives us:s=0+12·30·1Calculating further:s=0+12·30s=0+15Thus, the rocket will rise to a height of 15 meters.

A water rocket, launched from the ground, rises verticallywith acceleration of 30 /s2 for 1 s when it runs out of fuel. Disregarding air resistance, how high will the rocket rise?Comments

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