Question

After 1 second, the ball is 121 feet in the air; after 2 seconds, it is 224 feet in the air.Find the height, in feet, of the ball after 3 seconds in the air.

Answer 1

Answer:

The height, in feet, of the ball after 3 seconds in the air = 309

Step-by-step explanation:

We have $s = ut +\frac{1}{2}at^2$

After 1 second, the ball is 121 feet in the air

     Here s = 121 ft ,  t = 1 s

    $121=u\times 1-\frac{1}{2}\times a\times 1^2\\\\2u-a=242$

After 2 seconds, it is 224 feet in the air  

     Here s = 224 ft ,  t = 2 s  

     $224=u\times 2-\frac{1}{2}\times a\times 2^2\\\\u-a=112$

Subtracting two equations

         2u-a-(u-a)=242-112

          u = 130 ft/s

          130 - a = 112

            a = 18 ft/s²

When t = 3s

          $s=130\times 3-\frac{1}{2}\times 18\times 3^2=309ft$

The height, in feet, of the ball after 3 seconds in the air = 309