Question

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Explanation:

Step1: Understand the problem

We need to find when the ball hits the ground, which means the height \( y = 0 \). So we set the equation \( -16t^{2}+75t = 0 \).

Step2: Factor the equation

Factor out a \( t \) from the left - hand side of the equation: \( t(-16t + 75)=0 \).

Step3: Solve for t

According to the zero - product property, if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).

• Case 1: \( t = 0 \). This is the time when the ball is kicked (initial time), so we are interested in the other solution.
• Case 2: \( -16t+75 = 0 \). Solve for \( t \):
• First, subtract 75 from both sides: \( -16t=-75 \).
• Then, divide both sides by - 16: \( t=\frac{75}{16}=4.6875 \) seconds.

We are finding a "zero" because we are looking for the time when the height \( y = 0 \) (the ball hits the ground), which corresponds to the roots (zeros) of the quadratic equation \( y=-16t^{2}+75t \).

Answer:

Are you finding a "zero" OR "vertex": zero