The height of the ball after 2 seconds is 36 units
Solution:
The pathway of the ball from the machine can be represented by the equation:
[tex]h = -16t^2 + 48t + 4[/tex]
To find: height of the ball after 2 seconds
Substitute t = 2 in above equation
[tex]h = -16(2)^2 + 48(2) + 4\\\\h = -16 \times 4 + 96 + 4\\\\h = -64 + 100\\\\h = 36[/tex]
Thus the height of the ball after 2 seconds is 36 units
