While on a boat tour along the coast of the Atlantic Ocean, Leo sees a great white shark jump out of the water. The shark travels upward with a velocity of 18 feet per second before splashing back down into the water.To the nearest tenth of a second, how long is the shark in the air?Hint: Use the formula h=–16t2+vt+s. secondsSubmit
Question
While on a boat tour along the coast of the Atlantic Ocean, Leo sees a great white shark jump out of the water. The shark travels upward with a velocity of 18 feet per second before splashing back down into the water.To the nearest tenth of a second, how long is the shark in the air?Hint: Use the formula h=–16t2+vt+s. secondsSubmit
Solution
The formula given is a quadratic equation for the height (h) of the shark at any time (t), where v is the initial velocity and s is the initial height. In this case, v = 18 feet per second and s = 0 (since the shark starts at the surface of the water). The shark is in the air from the time it leaves the water until it returns to the water, which is when its height is 0. So, we need to solve the equation for t when h = 0.
0 = -16t^2 + 18t + 0
This is a quadratic equation in the form of at^2 + bt + c = 0, where a = -16, b = 18, and c = 0. We can solve for t using the quadratic formula, t = [-b ± sqrt(b^2 - 4ac)] / (2a).
t = [-18 ± sqrt((18)^2 - 4*-160)] / (2-16) t = [-18 ± sqrt(324)] / -32 t = [-18 ± 18] / -32
This gives us two solutions: t = 0 seconds (when the shark first jumps out of the water) and t = 1.125 seconds (when the shark lands back in the water). So, the shark is in the air for approximately 1.1 seconds, to the nearest tenth of a second.
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