Answer:
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;
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] - 2gs
where the variables have there usual meaning.
[tex]V^{2}[/tex] = 0 m/s, [tex]U^{2}[/tex] = 6 m/s, g = 9.8 m/[tex]s^{2}[/tex].
0 = [tex]6^{2}[/tex] - 2 x 9.8 x s
0 = 36 - 19.6s
⇒ 19.6s = 36
s = [tex]\frac{36}{19.6}[/tex]
= 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 = [tex]\frac{2USin A}{g}[/tex]
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 = [tex]90^{o}[/tex] (since the ball is thrown vertically upward), and g = 9.8 m/[tex]s^{2}[/tex].
T = [tex]\frac{2 *6 Sin90^{o} }{9.8}[/tex]
= [tex]\frac{12}{9.8}[/tex]
= 1.2245
T = 1.23s
The ball's time in the air is 1.23s.
