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

answer: point c: b. 1/2 potential, 1/2 kinetic, point a: 3/4 of its potential energy, point d or b: 3/4 potential, 1/4 kinetic

explanation: i cant explain. it's difficult to explain.