A tree frog is sitting on a leaf 4 feet above the ground. It jumps up with a velocity of 3 feet per second and lands on the ground below.To the nearest tenth of a second, how long is the frog in the air?Hint: Use the formula h=–16t2+vt+s.
Question
A tree frog is sitting on a leaf 4 feet above the ground. It jumps up with a velocity of 3 feet per second and lands on the ground below.To the nearest tenth of a second, how long is the frog in the air?Hint: Use the formula h=–16t2+vt+s.
Solution
Sure, let's solve this step by step.
The formula given is a quadratic equation that describes the height (h) of the frog at any time (t). The formula is h = -16t^2 + vt + s, where:
- h is the height at time t,
- v is the initial velocity (in feet per second),
- s is the initial height (in feet), and
- t is the time (in seconds).
In this case, the frog starts 4 feet above the ground (s = 4), jumps up with a velocity of 3 feet per second (v = 3), and we want to find when it lands on the ground (h = 0).
So, we substitute these values into the equation and solve for t:
0 = -16t^2 + 3t + 4.
This is a quadratic equation in the form of at^2 + bt + c = 0, where a = -16, b = 3, and c = 4. We can solve this equation for t using the quadratic formula, which is t = [-b ± sqrt(b^2 - 4ac)] / (2a).
Substituting the values of a, b, and c into the quadratic formula gives:
t = [-3 ± sqrt((3)^2 - 4*(-16)4)] / (2-16) = [-3 ± sqrt(9 + 256)] / -32 = [-3 ± sqrt(265)] / -32.
We get two solutions for t, one with the plus sign and one with the minus sign. However, time cannot be negative, so we discard the solution with the minus sign.
So, t = [-3 + sqrt(265)] / -32 ≈ 0.6 seconds.
Therefore, to the nearest tenth of a second, the frog is in the air for approximately 0.6 seconds.
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