Answer: The ball is <em><u>50 m</u></em> off the ground after 2 seconds
Step-by-step explanation:
Given the function relating the height of an object off the ground to the time spent falling is a quadratic relationship.
Therefore if h=height and t=time then
----------(A)
where a,b and c are constants
Apply given conditions
At t=0s h=90 m
=> 90 m = a+0+0
=>a=90 m
Also the ball has been just dropped at t=0 s
=>
=>
For t=0s b = 0
Thus equation (A) is reduced to 
At t= 3 s , h=0 m

Finally we get 
Therefore at t= 2.0 s , 
Thus the ball is <em><u>50 m</u></em> off the ground after 2 seconds