An archer shoots an arrow horizontally that hits a target 100m away. a bystander times the flight of the arrow at 2.2s. a) what was the initial velocity of the arrow? b) how far did the arrow drop in the vertical direction?

The archer is fired horizontally.

It hits the target which is at 100 m distance.

The horizontal velocity of the archer is always constant for a projectile in this situation.

The total time of flight is given as 2.2 s

Let the initial velocity is u .

Hence the target distance = ut

                 ⇒ 100 m = u×2.2 s

                   ⇒u = 100÷2.2 m/s

                           =45.45 m/s

We are asked to calculate the vertical distance travelled.

The vertical distance travelled is calculated as-

                                       [ s is the distance and a is the acceleration]

                                      [g is te acceleration due to gravity]

                                       [here we have taken only magnitude]

                                             =23.716 m       [ans]

  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