Question
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
Answers
GPT-4o mini
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.
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.