Atoy rocket is launched with an initial ve- locity of 12.0 m> s in the horizontal direction from the roof of a 30.0-m-tall building. the rocket’s engine produces a horizontal acceleration of 11.60 m> s 3 2t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g, down- ward. ignore air resistance. what horizontal distance does the rocket travel before reaching the ground?

D=90.3m.

Explanation:

Here we got a parabolic motion problem, On the Y axis we have a free-fall motion, so the time to reach the ground is given by:

Having that time we now can calculate the horizontal distance traveled by the rocket.

The acceleration is a continous function, if we integrate the acceleration we obtain the velocity:

the displacement is given by:

Substituting the time:

D=90.3 m

  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

it is a negative acceleration