Afisherman casts his lure at an angle of 33 degrees above the horizontal. the lure reaches a maximum height of 2.3 m. assuming no frictional forces, what was the initial velocity the fisherman gave the lure when he cast it?
(1 point)

12.34 m/s

28.45 m/s

34.91 m/s

21.29 m/s

The initial velocity is 12.34[m/s]

Explanation:

To understand this problem, let's draw a sketch of what is happening. In the attached image we can fin de sketch.

According to the equations of uniform movement accelerated we have to express them in the two components x and y.

In this problem we will analyze first the movement in the y axis and its equation to obtain the velocity in the initial point (o) and then with the angle of 33° respect to the horizontal we can find the main velocity.

In the previous equation we have the negative term - (2*a*y), the reason for the negative sign is because gravity is the acceleration that is acting downwards.

Therefore:

Now with the velocity in the y component, we can calculate the main vector of the velocity.

The main topic to understand in this problem is that when the height is maximum the velocity will be zero in the y component, therefore Vy=0 and we can simplify the problem.

  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

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more