Answer :
4.5 seconds it remain in the air after it remain in the air after thrown straight up with an initial velocity of 22.5 m/s
u initial velocity, final velocity v, distance s and time taken t.
Now, u = 22.5 m/s
v = 0 (at highest point)
g(during upward motion) = -10 m/s^2
Using kinematics formula:
[tex]v^2-u^2=2gs[/tex]
0-(22.5)^(2)=2×-10 × s
=> - 506.25 = -20 s
=> s = 506.25/20
=> s = 25.3125 m
v = u +gt
=> 0 = 22.5 -10 t
=> t = 22.5/10
=> t = 2.25 seconds
Total time the ball remains in air
= Time taken in going up + Time taken in coming down
= 2.25 + 2.25
= 4.5 seconds
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