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Chapter 3: Problem 92

A gull is flying horizontally \(8.00 \mathrm{m}\) above the ground at $6.00 \mathrm{m} / \mathrm{s} .$ The bird is carrying a clam in its beak and plans to crack the clamshell by dropping it on some rocks below. Ignoring air resistance, (a) what is the horizontal distance to the rocks at the moment that the gull should let go of the clam? (b) With what speed relative to the rocks does the clam smash into the rocks? (c) With what speed relative to the gull does the clam smash into the rocks?

Short Answer

Expert verified
a) The horizontal distance to the rocks when the gull should drop the clam is approximately 7.68 meters. b) The speed of the clam relative to the rocks upon impact is approximately 13.86 m/s. c) The speed of the clam relative to the gull upon impact is approximately 12.56 m/s.

Step by step solution

01

Determine time it takes for the clam to hit the ground#End#

First we need to know how long it takes for the clam to fall from the height of 8 meters. Since the motion is vertical and acceleration due to gravity always acts downward, we can use the following kinematic equation: \(h = v_{0y} t + \frac{1}{2} a_y t^2\) where \(h = 8\) m (initial vertical position of the clam) \(v_{0y} = 0\) m/s (clam is dropped from rest) \(a_y = -9.81\) m/s² (acceleration due to gravity) Solving for t, we get: \(8 = 0 * t + \frac{1}{2} * (-9.81) * t^2\)
02

Calculate time#End#

To solve for t, we can rearrange the equation as a quadratic equation: \(t^2 = \frac{2 * 8}{9.81}\) Now, solve for t: \(t = \sqrt{\frac{2 * 8}{9.81}}\) Calculating this value yields: \(t \approx 1.28\) s.
03

Calculate horizontal distance to the rocks (Answer to part a) #End#

Since the gull is flying horizontally, we can use the following equation to determine the horizontal distance x, the gull should drop the clam: \(x = v_x * t\) Where \(v_x = 6\) m/s (horizontal velocity of the gull) Plugging in the values, we have: \(x = 6 * 1.28\) Calculating this value yields: \(x \approx 7.68\) m. Thus, the horizontal distance to the rocks at the moment the gull should let go of the clam is approximately 7.68 meters.
04

Calculate the final vertical velocity of the clam (Before impact) #End#

To find the final vertical velocity \(v_y\) of the clam just before impact, we can use the equation: \(v_y = v_{0y} + a_y * t\) Where \(v_{0y} = 0\) m/s \(a_y = -9.81\) m/s² \(t \approx 1.28\) s Plug in the values, and we have: \(v_y = 0 + (-9.81) * 1.28\) Calculating this value yields: \(v_y \approx -12.56\) m/s.
05

Calculate speed relative to the rocks (Answer to part b) #End#

To compute the speed relative to the rocks, we need to find the magnitude of the velocity vector: \(v_{clam} = \sqrt{v_x^2 + v_y^2}\) Where \(v_x = 6\) m/s \(v_y \approx -12.56\) m/s Plug in the values, and we have: \(v_{clam} = \sqrt{(6)^2 + (-12.56)^2}\) Calculating this value yields: \(v_{clam} \approx 13.86\) m/s. Therefore, the speed of the clam relative to the rocks is approximately 13.86 m/s.
06

Calculate speed relative to the gull (Answer to part c) #End#

Since the gull is flying horizontally at a constant velocity, the relative speed of the clam to the gull is the same as the vertical speed of the clam before impact: \(v_{relative} = v_y \approx -12.56\) m/s However, since we are looking for the magnitude of the relative velocity, we take the absolute value: \(v_{relative} = |v_y| \approx 12.56\) m/s. Thus, the speed of the clam relative to the gull is approximately 12.56 m/s.

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