You are standing on the shore by a creek skipping rocks across the water. The rock skips off the water at an angle of 6.00° and at a speed of 2.00 m/s. How far does the rock travel before it next hits the water? 0.0849 m 0.812 m 0.00892 m 4.87 m
Question
You are standing on the shore by a creek skipping rocks across the water. The rock skips off the water at an angle of 6.00° and at a speed of 2.00 m/s. How far does the rock travel before it next hits the water? 0.0849 m 0.812 m 0.00892 m 4.87 m
Solution
To solve this problem, we need to use the equation for the range of a projectile, which is given by:
R = (v²/g) * sin(2θ)
where:
- R is the range, or the horizontal distance the rock travels,
- v is the initial velocity of the rock,
- g is the acceleration due to gravity (approximately 9.81 m/s² on Earth), and
- θ is the launch angle.
Given that v = 2.00 m/s, g = 9.81 m/s², and θ = 6.00°, we can substitute these values into the equation:
R = (2.00 m/s)² / 9.81 m/s² * sin(2 * 6.00°) R = 4.00 m²/s² / 9.81 m/s² * sin(12.00°) R = 0.4076 m * sin(12.00°) R = 0.4076 m * 0.2079 R = 0.0847 m
So, the rock travels approximately 0.0847 meters, or 8.47 cm, before it next hits the water. The closest answer to this is 0.0849 m.
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