Abaseball is hit almost straight up into the air with a speed of 22 m/s .
estimate how high it goes
estimate how long its in the air

The maximum height at which the baseball will reach is given by the formula 

Vf^2 - Vo^2 = 2gh 

where 

Vf = final velocity = 0 (at its peak) 
Vo = initial velocity = 22 m/sec (given) 
g = acceleration due to gravity = 9.8 m/sec^2 (constant) 
h = maximum height attained by the ball 

Substituting appropriate values, 

0 - (22)^2 = 9(-9.8)h 

NOTE the negative sign attached to the acceleration due to gravity. This simply denotes that the ball is slowing down as it goes up. 

Solving for "h", 

h = 22^2/(2 * 9.8) 

h = 24.69 meters 

To solve for the time it reaches its maximum height, use the formula 

h = VoT -(1/2)(gT^2) 

where 

T = time for ball to reach maximum height 

and all the other terms have been previously defined. 

Substituting values, 

24.69 = 22T - (1/2)(9.8)T^2 

Rearranging the above, 

4.9T^2 - 22T + 24.69 = 0 

Using the quadratic formula, 

T = 2.22 seconds 

NOTE --- the above is the time for the ball to reach its maximum height. The total time it is in the air is 

Total time = 2(2.22) = 4.44 sec.

Hope this helps :) 

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