Question

algebra skill review #2
score: 12/21 answered: 12/21
question 13
if an object is thrown straight up into the air with an initial velocity of 32 feet per second, then its height above the ground at any time t is given by the formula ( h = 32t - 16t^{2} ). find the times at which the object is on the ground by letting ( h = 0 ) in the equation and solving for t.
( t=square,square ) seconds
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Explanation:

Step1: Substitute \( h = 0 \) into the equation

Substitute \( h = 0 \) into \( h=32t - 16t^{2} \), we get \( 0 = 32t-16t^{2} \).

Step2: Factor out the common term

Factor out \( 16t \) from the right - hand side of the equation \( 0 = 32t-16t^{2} \). So, \( 0=16t(2 - t) \).

Step3: Use the zero - product property

According to the zero - product property, if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).
For \( 16t(2 - t)=0 \), we have two cases:

• Case 1: \( 16t=0 \), then \( t = 0 \).
• Case 2: \( 2 - t=0 \), then \( t = 2 \).

Answer:

\( t = 0,2 \) seconds