Tim launches an object at an upward velocity of 49feet per second taking the formula y = -16t^2 +40 t - 5 How are far above or below is the initial launching spot

Answer 1

To determine how far above or below the initial launching spot the object is at time \( t = 0 \), we can evaluate the function \( y(t) \) at \( t = 0 \).

The formula provided is:

\[ y = -16t^2 + 40t - 5 \]

Substituting \( t = 0 \):

\[ y(0) = -16(0)^2 + 40(0) - 5 \] \[ y(0) = -5 \]

Thus, at \( t = 0 \), the position \( y \) of the object is \(-5\) feet. This means the object is 5 feet below the initial launching spot.