A ball is thrown with an initial upward velocity of 5 m/s.

A ball is thrown at an initial height of 5 feet with an initial upward velocity at 29 ft/s. lets assume that  balls height h (in feet) after t seconds is give by:

h= 5 + 29t -16t^2

Explanation:

h= 5 + 29t -16t^2

a time when the ball's height will be 17 ft

17 = 5 + 29t -16t2

0 = -17 + 5 + 29t -16t2

0 = -12 + 29t - 16t2

Using the quadratic equation:

t = (-29±√(292-(4*(-16)*(-12÷2(-16)

 = (-29±√(841 - 768))÷(-32)

 = (-29±√(73))÷(-32)

 = (-29 + 8.544)÷(-32)   or   (-29 - 8.544)÷(-32)

 = (-20.456)÷(-32)         or   -37.544÷(-32)

 =  0.64                         or   1.17

So, the ball is at a height of 17 ft twice: once on the way up after 0.64 seconds and once on the way back down after 1.17 seconds.

C2H6O

Explanation:

Empirical formula is the formula showing proportions of the elements present in a compound. Empirical formula will not show the arrangements of atoms in a compound.

From the question above, we are given that the compound contains 52.14% of carbon, 34.73% of oxygen and the amount of Hydrogen is unknown.

Therefore, to determine the amount of Hydrogen, we say;

100% of the whole compound - (52.14% + 34.73%).

Percentage of hydrogen in the compound= 13.13% .

Therefore, Mole of Hydrogen atom= 13.13/1 = 13.13.

Mole of carbon atom = 52.14/12= 4.345.

Mole of Oxygen atom= 34.73/16 = 2.171.

The next thing to do is to divide the each mole by the smallest mole(that is, the mole of oxygen atom).

Hence, Hydrogen,H = 13.13/2.171 =6.

Carbon,C= 4.345/2.171 = 2.

Oxygen = 2.171/2.171 = 1.

Therefore the empirical formula is;

C2H6O.