Arocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. the rocket moves for 3.00 s along its initial line of motion with an acceleration of 32.0 m/s2. at this time, its engines fail and the rocket proceeds to move as a projectile. (a) find the maximum altitude reached by the rocket. 102 incorrect: your answer is incorrect. your response differs significantly from the correct answer. rework your solution from the beginning and check each step carefully. m (b) find its total time of flight. s (c) find its horizontal range. 186 incorrect: your answer is incorrect. your response differs significantly from the correct answer. rework your solution from the beginning and check each step carefully. m

Before the engines fail , the rocket's horizontal and vertical position in the air are

and its velocity vector has components

After , its position is

and the rocket's velocity vector has horizontal and vertical components

After the engine failure , the rocket is in freefall and its position is given by

and its velocity vector's components are

where we take .

a. The maximum altitude occurs at the point during which :

At this point, the rocket has an altitude of

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve for , then add 3 seconds to this time:

So the rocket stays in the air for a total of .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute for this time :

good looking response

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂