Afootball game begins with a coin toss to determine who kicks off. the referee tosses the coin upward with an initial speed of 5 m/s, ignore air resistance.
a)how high does the coin go above the point of release?
b)what is the total time the coin is in the air before returning to the release point?

(a) 1.28 m

(b) 0.51 s

Explanation:

Given:

Initial speed, U = 5 m/s

Consideration;

1. When the coin is tossed upward, the coin is moving against gravity. Therefor, the acceleration due to gravity experienced by the coin is negative, -g = -9.8 m/s².

2. The final velocity of the coin in the air is zero.  

From equation of motion under free fall,

                           V² = U² + 2gh

                            h =

Where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

h is the height

                              h = 1.2755 m

                              h = 1.28 m  

(b) The coin still has its initial velocity, u to be equal to 5 m/s, final velocity, v to be equal to 0 m/s and acceleration due to gravity to be -9.8 m/s while in the air before returning to the release point.

From equation of motion under free fall,

                                     V = U + gt

                                     t = 0.51 s

  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