Questions: A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s(t)=112t-16t^2. After how many seconds will the ball be 160 feet from the ground? (Hint: Look for a common factor before solving the equation.) seconds (Use a comma to separate answers as needed.)

A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s(t)=112t-16t^2. After how many seconds will the ball be 160 feet from the ground? (Hint: Look for a common factor before solving the equation.)
seconds
(Use a comma to separate answers as needed.)

Transcript text: A ball is thrown vertically upward from the ground. Its distance in feet from the ground in $t$ seconds is $s(t)=112 t-16 t^{2}$. After how many seconds will the ball be 160 feet from the ground? (Hint: Look for a common factor before solving the equation.) $\square$ seconds (Use a comma to separate answers as needed.)

Solution

Solution Steps

Step 1: Set Up the Equation

To find when the ball is 160 feet from the ground, set the distance function equal to 160: \[ s(t) = 112t - 16t^2 = 160 \]

Step 2: Rearrange the Equation

Rearrange the equation to bring all terms to one side: \[ 112t - 16t^2 - 160 = 0 \]

Step 3: Factor the Equation

Factor out the greatest common factor from the equation: \[ -16(t^2 - 7t + 10) = 0 \]

Step 4: Solve the Quadratic Equation

Solve the quadratic equation \( t^2 - 7t + 10 = 0 \) by factoring: \[ (t - 5)(t - 2) = 0 \]

Step 5: Find the Solutions

Set each factor equal to zero and solve for \( t \): \[ t - 5 = 0 \quad \Rightarrow \quad t = 5 \] \[ t - 2 = 0 \quad \Rightarrow \quad t = 2 \]

Step 6: List the Solutions

The ball will be 160 feet from the ground at \( t = 2 \) seconds and \( t = 5 \) seconds.

Final Answer

\(\boxed{2, 5}\)

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