Question

A particle is fired straight up from the ground. Its height in feet after \(t\) second is given by \(s(t) = 128t-16t^2\). The velocity of the particle when it hits the ground is...

• A. \(-128\) ft/sec
• B. \(128\) ft/sec
• C. \(0\) ft/sec
• D. \(256\) ft/sec

Hint:

Find the time when s = 0 again, then differentiate s(t).

Answer 1

The Correct Option is A

Step 1: Symmetry:
Motion under constant gravity is symmetric: the particle lands with the same speed it started with, but in the opposite direction. The launch velocity is $v(0)=128$.

Step 2: Landing Velocity:
So the landing velocity is $-128$ ft/s. Check: $v(8)=128-32\cdot8=-128$.

Step 3: Answer:
Option (A).

Final Answer:
Option (A). \[ \boxed{\text{(A) } -128\ \text{ft/s}} \]