Question

A ball was tossed up into the air. The height of the ball as a function of the time the ball is in the air in seconds can be modeled by a quadratic function. At the time the ball was initially thrown in the air, it was 7. 5 meters off the ground. After 1. 25 seconds, the ball was at its maximum height of 15. 3125 meters. What was the height of the ball 2 seconds after being thrown? Round your answer to the nearest tenth as needed. Meters.

Answer 1

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;

$m=\frac{15.3125-7.5}{1.25-0}\\m= \frac{7.8125}{1.25}\\m= 6.25$

Get the y-intercept "b"

Recall that y = mx + b

$7.5=6.25(0) + b\\7.5 = b\\ b= 7.5$

Get the required equation:

$y = 6.65x + 7.5$

Next is to get the height of the ball 2 seconds after being thrown

$y = 6.65x + 7.5\\y = 6.65(2)+ 7.5\\y=13.3 + 7.5\\y=20.8m$

Hence the height of the ball 2 seconds after being thrown is 20.8m

Learn more on functions here: brainly.com/question/1214333