The height of the ball 2 seconds after being thrown is 20.8m
In order to get the required quadratic function, we need to write the given information as coordinate.
- If the time the ball was initially thrown in the air, it was 7. 5 meters off the ground, this is written in a coordinate form as (0, 7.5)
- If after 1. 25 seconds, the ball was at its maximum height of 15.3125 meters in a coordinate form as (1.25, 15.3125)
Get the slope of the equation;
[tex]m=\frac{15.3125-7.5}{1.25-0}\\m= \frac{7.8125}{1.25}\\m= 6.25[/tex]
Get the y-intercept "b"
Recall that y = mx + b
[tex]7.5=6.25(0) + b\\7.5 = b\\ b= 7.5[/tex]
Get the required equation:
[tex]y = 6.65x + 7.5[/tex]
Next is to get the height of the ball 2 seconds after being thrown
[tex]y = 6.65x + 7.5\\y = 6.65(2)+ 7.5\\y=13.3 + 7.5\\y=20.8m[/tex]
Hence the height of the ball 2 seconds after being thrown is 20.8m
Learn more on functions here: https://brainly.com/question/1214333
