/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 89 A locust jumps at an angle of \(... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 89

A locust jumps at an angle of \(55.0^{\circ}\) and lands \(0.800 \mathrm{m}\) from where it jumped. (a) What is the maximum height of the locust during its jump? Ignore air resistance. (b) If it jumps with the same initial speed at an angle of \(45.0^{\circ},\) would the maximum height be larger or smaller? (c) What about the range? (d) Calculate the maximum height and range for this angle.

Short Answer

Expert verified
Answer: To determine whether the locust would achieve a higher maximum height and range for a \(45.0^{\circ}\) angle jump than for a \(55.0^{\circ}\) angle jump, compare the values calculated in Step 4 and Step 5 for the maximum height and range, respectively.

Step by step solution

01

Calculate the initial horizontal velocity

In order to find the initial horizontal velocity, we can use the range formula: \(R=\frac{v_{0x}t}{2}\), where \(R\) is the range, \(v_{0x}\) is the initial horizontal velocity, and \(t\) is the time for which locust is in the air. Given \(R=0.800\)m and considering that the horizontal velocity remains constant, we have: \(v_{0x} = \frac{2R}{t}\)
02

Calculate the initial total velocity

Now we will calculate the initial total velocity using the initial horizontal velocity we just found and the angle with which the locust jumped (in radians). We can convert the given angle from degrees to radians: \(\theta = 55.0^{\circ}\times\frac{\pi}{180}=0.9599\) radians Using the relationship: \(v_{0x}=v_0\cos{\theta}\), we then solve for \(v_0\): \(v_0 = \frac{v_{0x}}{\cos{\theta}}\)
03

Calculate the initial vertical velocity

Now we will calculate the initial vertical velocity in order to find the maximum height. We will use the following relationship: \(v_{0y} = v_0\sin\theta\)
04

Determine the maximum height

To find the maximum height, we can use the following kinematic equation considering that the final vertical velocity is zero at the maximum height: \(v_y^{2}=v_{0y}^{2}- 2gh\) Solving for \(h\): \(h = \frac{v_{0y}^2}{2g}\)
05

Calculate the maximum height and range for a \(45.0^{\circ}\) angle jump

Now that we have found the initial total velocity, we can calculate the height and range when the jumping angle is \(45.0^{\circ}\). First, we need to convert the angle to radians: \(\theta = 45.0^{\circ}\times\frac{\pi}{180}=0.7854\) radians. We can find the initial horizontal and vertical velocities using: \(v_{0x} = v_0\cos{\theta}\) \(v_{0y} = v_0\sin{\theta}\) The maximum height formula remains the same as in step 4. For the range, we can use the same formula as in step 1 with the help of the new horizontal velocity and time in the air.
06

Compare the maximum height and range for both angles

Now to answer part (b) and (c), we can compare the heights and ranges obtained from step 4 and 5, respectively. If the height from step 5 is greater than step 4's height, then the jump would have a larger maximum height. Similarly, if the range from step 5 is greater than step 1's range, then the jump would have a larger range.

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