Questions: An arrow is shot vertically upward at a rate of 220 ft / s. Use the projectile formula h=-16 t^2+v0 t to determine at what time(s), in seconds, the arrow is at a height of 400 ft. Round your answer(s) to the nearest tenth of a second.

An arrow is shot vertically upward at a rate of 220 ft / s. Use the projectile formula h=-16 t^2+v0 t to determine at what time(s), in seconds, the arrow is at a height of 400 ft. Round your answer(s) to the nearest tenth of a second.

Transcript text: Question An arrow is shot vertically upward at a rate of $220 \mathrm{ft} / \mathrm{s}$. Use the projectile formula $h=-16 t^{2}+v_{0} t$ to determine at what time(s), in seconds, the arrow is at a height of 400 ft . Round your answer(s) to the nearest tenth of a second.

Solution

Solution Steps

Step 1: Rearrange the equation to standard quadratic form

The equation is rearranged to \(0 = -16t^2 + v_0t - h\).

Step 2: Apply the quadratic formula

The quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is applied with \(a = -16\), \(b = v_0\), and \(c = -h\).

Step 3: Calculate the discriminant

The discriminant \(D = b^2 - 4ac\) is calculated as \(D = 22800\). Since \(D > 0\), there are two distinct real solutions.

Step 4: Solve for \(t\)

The solutions are \(t_1 = 2.2\) and \(t_2 = 11.6\).

Final Answer:

The projectile reaches the height \(h\) at two different times: \(t_1 = 2.2\) seconds and \(t_2 = 11.6\) seconds.

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