Zack is on his school's color guard and is practicing throwing and catching his flag. He throws the flag into the air from a height of 5 feet at a velocity of 30 feet per second. After the flag starts to come back down, he catches it 7 feet above the ground.To the nearest tenth of a second, how long is the flag in the air before Zack catches it?Hint: Use the formula h=–16t2+vt+s.
Question
Zack is on his school's color guard and is practicing throwing and catching his flag. He throws the flag into the air from a height of 5 feet at a velocity of 30 feet per second. After the flag starts to come back down, he catches it 7 feet above the ground.To the nearest tenth of a second, how long is the flag in the air before Zack catches it?Hint: Use the formula h=–16t2+vt+s.
Solution
To solve this problem, we will use the formula for the height of a projectile in free fall: h = -16t^2 + vt + s. In this formula, h is the height of the object, t is the time it has been in the air, v is the initial upward velocity, and s is the initial height.
In this case, we know that Zack catches the flag 7 feet above the ground (h), he throws it with an initial velocity of 30 feet per second (v), and he throws it from a height of 5 feet (s). We want to find the time it takes for the flag to reach Zack's hand (t).
So, we plug these values into the formula:
7 = -16t^2 + 30t + 5
We can simplify this equation by subtracting 7 from both sides to set it equal to zero:
0 = -16t^2 + 30t - 2
This is a quadratic equation, and we can solve it using the quadratic formula: t = [-b ± sqrt(b^2 - 4ac)] / (2a). In this case, a = -16, b = 30, and c = -2.
t = [-30 ± sqrt((30)^2 - 4*(-16)(-2))] / (2(-16)) t = [-30 ± sqrt(900 - 128)] / -32 t = [-30 ± sqrt(772)] / -32
This gives us two possible solutions: t = 0.6 seconds and t = 2.1 seconds. However, since the flag cannot be in the air for a negative amount of time, we discard the negative solution.
Therefore, Zack catches the flag approximately 2.1 seconds after he throws it.
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