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: Find the Landing Time:
The particle is on the ground when \(s=0\): \(128t-16t^2=16t(8-t)=0\). So \(t=0\) (launch) or \(t=8\) s (landing).

Step 2: Velocity Function:
\(v=\dfrac{ds}{dt}=128-32t\).

Step 3: Evaluate at Landing:
\(v(8)=128-32(8)=128-256=-128\ \text{ft/s}\).

Step 4: Meaning:
The negative sign shows the particle moves downward at 128 ft/s when it hits the ground. The speed equals the launch speed, so the magnitude is 128. Option (B) has the wrong sign, (C) is the velocity at the peak, and (D) is unrelated.

Final Answer:
The velocity is \(-128\) ft/s, option (A). \[ \boxed{\text{(A) } -128\ \text{ft/s}} \]