While warming up at soccer practice, Ellie's teammate tosses the ball to Ellie. Ellie hits the ball upward with her head from a height of 1.6 meters at a velocity of 6 meters per second. The ball is in the air for some time until Ellie catches it at a height of 1 meter.To the nearest tenth of a second, how long is the ball in the air before Ellie catches it?Hint: Use the formula h=–4.9t2+vt+s. seconds
Question
While warming up at soccer practice, Ellie's teammate tosses the ball to Ellie. Ellie hits the ball upward with her head from a height of 1.6 meters at a velocity of 6 meters per second. The ball is in the air for some time until Ellie catches it at a height of 1 meter.To the nearest tenth of a second, how long is the ball in the air before Ellie catches it?Hint: Use the formula h=–4.9t2+vt+s. seconds
Solution
To solve this problem, we will use the formula h = -4.9t^2 + vt + s, where:
- h is the final height (1 meter in this case)
- t is the time in seconds (which we are trying to find)
- v is the initial velocity (6 meters per second in this case)
- s is the initial height (1.6 meters in this case)
We can plug these values into the formula and solve for t:
1 = -4.9t^2 + 6t + 1.6
Rearranging the equation to set it equal to zero gives us:
0 = -4.9t^2 + 6t + 0.6
This is a quadratic equation in the form of 0 = at^2 + bt + c. We can solve for t using the quadratic formula, t = [-b ± sqrt(b^2 - 4ac)] / (2a):
t = [-6 ± sqrt((6)^2 - 4*(-4.9)0.6)] / (2-4.9)
Solving this gives us two possible solutions for t. However, since time cannot be negative, we discard the negative solution and keep the positive one.
After calculating, we find that the ball is in the air for approximately 0.7 seconds before Ellie catches it.
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