Answer: Â The ball is 50 m 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
[tex]h=a+bt+ct^{2}[/tex] Â Â Â ----------(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
=>[tex]\frac{\partial h}{\partial t}=0=>\frac{\partial (a+bt+ct^{2})}{\partial t}=0[/tex]
=>[tex]b+2ct=0[/tex]
For t=0s b = 0
Thus equation (A) is reduced to [tex]h=90+ct^{2}[/tex]
At  t= 3 s , h=0 m
[tex]\therefore 0= 90 +9c=>c=-10 \frac{m}{s^{2}}[/tex]
Finally we get [tex]h=90-10t^{2}[/tex]
Therefore at t= 2.0 s , [tex]h=(90-10\times 2^{2})m=50 m[/tex]
Thus the ball is 50 m off the ground after 2 seconds
