Answer: The ball is 50 m off the ground after 2 seconds
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
=>
![(\partial h)/(\partial t)=0=>(\partial (a+bt+ct^(2)))/(\partial t)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hfen9fiymke5t09e6728jvsybh8wge1ys0.png)
=>
![b+2ct=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/3g93m5har3uybffsakqz3ohx6jtwb1xdvf.png)
For t=0s b = 0
Thus equation (A) is reduced to
![h=90+ct^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/z9xmqgerkqjwjxsn756zwinjf0t6qov77u.png)
At t= 3 s , h=0 m
![\therefore 0= 90 +9c=>c=-10 (m)/(s^(2))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/kz3w1muldlahzjz7vmipkl5c6datmtvx43.png)
Finally we get
![h=90-10t^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/vea0j9n5iarbv5xh407l5i21azyemlu1us.png)
Therefore at t= 2.0 s ,
![h=(90-10* 2^(2))m=50 m](https://img.qammunity.org/2020/formulas/mathematics/middle-school/dmbfdo7xlhzc0qf8kekns4lk1zs3ui0tvy.png)
Thus the ball is 50 m off the ground after 2 seconds