A Boy has a ball and throws it straight up into the air with an initial velocity of 6 m/s. Calculate the maximum height of the ball and the time the ball is in the air.

i. 1.84 m

ii. The ball's time in the air is 1.23s.

Explanation:

The ball moves upward (against the gravitational pull) with an initial velocity of 6 m/s.

I. Applying the third equation of free fall, the maximum height is;

= - 2gs

where the variables have there usual meaning.

= 0 m/s, = 6 m/s, g = 9.8 m/.

0 = - 2 x 9.8 x s

0 = 36 - 19.6s

⇒ 19.6s = 36

s =

 = 1.8367

 = 1.84 m

The maximum height of the ball is 1.84 m.

II. The time the ball was in air can be determined by,

time of flight, T =

where U is the initial velocity of the ball, A is the angle of projection and g is the acceleration due to gravity.

U = 6 m/s, A = (since the ball is thrown vertically upward), and g = 9.8 m/.

T =

  =

  = 1.2245

T = 1.23s

The ball's time in the air is 1.23s.

answer: 0.20

explanation: this can only be the answew because all the other measuremnets are not close

the area between the graph and the t axis