Questions: If a ball is thrown directly upward with a velocity of 20 ft / s, its height (in feet) after t seconds is given by y=20t-16t^2. What is the maximum height attained by the ball? (Round your answer to the nearest whole number.)

If a ball is thrown directly upward with a velocity of 20 ft / s, its height (in feet) after t seconds is given by y=20t-16t^2. What is the maximum height attained by the ball? (Round your answer to the nearest whole number.)

Transcript text: If a ball is thrown directly upward with a velocity of $20 \mathrm{ft} / \mathrm{s}$, its height (in feet) after $t$ seconds is given by $y=20 t-16 t^{2}$. What is the maximum height attained by the ball? (Round your answer to the nearest whole number.)

Solution

Solution Steps

Step 1: Convert the given quadratic equation into the standard form

The given equation is in the form $y = vt - gt^2 + h_0$. By identifying $a = -g$, $b = v$, and $c = h_0$, we convert it into the standard form $y = ax^2 + bx + c$, where $a = -16$, $b = 20$, and $c = 0$.

Step 2: Use the formula for the vertex of a parabola to find the time at which the maximum height is attained

Using the formula $t = -\frac{b}{2a}$, we find the time to reach the maximum height as $t = \frac{v}{2g} = 0.625$ seconds.

Step 3: Substitute this time back into the original equation to find the maximum height

Substituting $t = 0.625$ back into the original equation, we find the maximum height as $y_{max} = \frac{v^2}{4g} + h_0 = 6$ units.

Final Answer:

The maximum height attained by the object is 6 units.

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