Questions: A stone is thrown straight up from the roof of a 240-ft building. The distance (in feet) of the stone from the ground at any time t (in seconds) is given by the following. h(t)=-16 t^2+32 t+240 When is the stone rising? 5<t<1 t<5 t<1 t>5 t>1 When is it falling? 1>t>5 t<5 t<1 t>1 t>5 If the stone were to miss the building, when would it hit the ground? t= sec

Solution Steps

To find when the stone is rising, we need to determine when the velocity of the stone is positive. The velocity is the derivative of the height function \( h(t) \).

The height function is given by: \[ h(t) = -16t^2 + 32t + 240 \]

The velocity \( v(t) \) is the derivative of \( h(t) \): \[ v(t) = \frac{d}{dt}(-16t^2 + 32t + 240) = -32t + 32 \]

Set the velocity greater than zero to find when the stone is rising: \[ -32t + 32 > 0 \]

Solving for \( t \): \[ -32t > -32 \\ t < 1 \]

Thus, the stone is rising when \( t < 1 \).

The stone is falling when the velocity is negative: \[ -32t + 32 < 0 \]

Solving for \( t \): \[ -32t < -32 \\ t > 1 \]

Thus, the stone is falling when \( t > 1 \).

The stone hits the ground when \( h(t) = 0 \). We solve the equation: \[ -16t^2 + 32t + 240 = 0 \]

This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where \( a = -16 \), \( b = 32 \), and \( c = 240 \). We use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substitute the values: \[ t = \frac{-32 \pm \sqrt{32^2 - 4(-16)(240)}}{2(-16)} \]

Calculate the discriminant: \[ 32^2 - 4(-16)(240) = 1024 + 15360 = 16384 \]

Calculate \( t \): \[ t = \frac{-32 \pm \sqrt{16384}}{-32} \]

\[ t = \frac{-32 \pm 128}{-32} \]

This gives two solutions: \[ t = \frac{-32 + 128}{-32} = \frac{96}{-32} = -3 \] \[ t = \frac{-32 - 128}{-32} = \frac{-160}{-32} = 5 \]

Since time cannot be negative, the stone hits the ground at \( t = 5 \) seconds.

Final Answer