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MHT CET202619 April 2026Morning ShiftMathematicsApplication of DerivativesActual

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...

Options

  1. A-128 ft/sec
  2. B128 ft/sec
  3. C0 ft/sec
  4. D256 ft/sec

Correct answer

A. -128 ft/sec

Step-by-step solution

The height of the particle is given by s(t) = 128t - 16t^2 . The particle hits the ground when s(t) = 0 . 128t - 16t^2 = 0 16t(8 - t) = 0 This gives t = 0 or t = 8 . The particle hits the ground at t = 8 seconds. The velocity of the particle is given by the derivative of the position function: v(t) = ds dt = 128 - 32t Substituting t = 8 into the velocity equation: v(8) = 128 - 32(8) = 128 - 256 = -128 ft/sec. Answer: -128 ft/sec

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