During a baseball game, a batter hits a popup to a fielder 83 m away. The acceleration of gravity is 9.8 m/s
2
.
If the ball remains in the air for 5.7 s, how
high does it rise?
Answer in units of m

The height risen by the ball is 39.8 m.

Explanation:

Given;

total time spent in air by the ball, t = 5.7 s

time to rise to maximum height, t = 5.7 /2 = 2.85 s

The initial velocity of the ball is given by;

v = u - gt

where;

v is the final velocity at maximum height = 0

u is the initial velocity of the ball

0 = u - gt

u = gt

u = (9.8 x 2.85)

u = 27.93 m/s

The vertical height traveled by the ball is given by;

v² = u² - 2gh

where;

v is the final velocity of the ball at maximum height = 0

g is acceleration due to gravity = 9.8 m/s²

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (27.93)² / ( 2 x 9.8)

h = 39.8 m

Therefore, the height risen by the ball is 39.8 m.

the answer is 3.1 his answer is wrong mines is right