COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 4, Problem 86P

(a)

To determine

Maximum height of the locust during the its jump.

(a)

Expert Solution
Check Mark

Answer to Problem 86P

Maximum height of the locust during the its jump is 0.28m_.

Explanation of Solution

Write the expression for horizontal range of the projectile motion.

    R=vi2sin2θg        (I)

Here, vi is the initial velocity, θ is the projectile angle, g is the acceleration due to gravity, and R is the horizontal range.

Solve equation (I) for vi.

    vi=Rgsin2θ        (II)

Write the kinematic equation.

    vyf2=vyi2+2ayy        (III)

Here, vyf is the final velocity in y direction, vyi is the initial velocity in y direction, y is the vertical distance, and ay is the acceleration in y direction.

Substitute, 0 for vyf, visinθ for vyi, and H for y in the equation (III), to find maximum height.

    0=(visinθ)22gHH=(visinθ)22g        (IV)

Conclusion:

Substitute, 0.8m for R, 55° for θ, and 9.8m/s2 for g in the equation (II), to find vi.

    vi=(0.8m)(9.8m/s2)sin2×55°=2.9m/s

Substitute, 2.9m/s for vi, 55° for θ, and 9.8m/s2 for g in the equation (IV), to find H1.

    H1=[(2.9m/s)(sin55°)]22(9.8m/s2)=0.28m

Therefore, the maximum height of the locust during the its jump is 0.28m_.

(b)

To determine

Whether the maximum height will be smaller or larger if θ=45°.

(b)

Expert Solution
Check Mark

Answer to Problem 86P

The maximum height will be smaller_ if θ=45°.

Explanation of Solution

Conclusion:

Substitute, 2.9m/s for vi, 45° for θ, and 9.8m/s2 for g in the equation (IV), to find H2.

    H2=[(2.9m/s)(sin45°)]22(9.8m/s2)=0.21m

Therefore, the maximum height will be smaller_ if θ=45°.

(c)

To determine

Whether the range will be smaller or larger if θ=45°.

(c)

Expert Solution
Check Mark

Answer to Problem 86P

The range will be larger_ if θ=45° than θ=55°.

Explanation of Solution

Conclusion:

Substitute, 2.9m/s for vi, 45° for θ, and 9.8m/s2 for g in the equation (I), to find R2.

    R2=(2.9m/s)2sin2(45°)9.8m/s2=0.585

Therefore, the range will be larger_ if θ=45° than θ=55°.

(d)

To determine

The maximum height and range if θ=45°.

(d)

Expert Solution
Check Mark

Answer to Problem 86P

The maximum height if θ=45° is 0.21m_, and maximum range is 0.858m.

Explanation of Solution

From the part (c), it is clear that the maximum height if θ=45° is 0.21m. And From the part (d), it is clear that the range if θ=45° is 0.858m.

Conclusion:

Therefore, The maximum height if θ=45° is 0.21m_, and maximum range is 0.858m_.

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Chapter 4 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

Ch. 4.4 - Prob. 4.8PPCh. 4.5 - Prob. 4.5ACPCh. 4.5 - Prob. 4.10PPCh. 4.5 - Prob. 4.5BCPCh. 4.6 - Prob. 4.11PPCh. 4.6 - Prob. 4.6CPCh. 4 - Prob. 1CQCh. 4 - Prob. 2CQCh. 4 - Prob. 3CQCh. 4 - Prob. 4CQCh. 4 - Prob. 5CQCh. 4 - Prob. 6CQCh. 4 - Prob. 7CQCh. 4 - Prob. 8CQCh. 4 - Prob. 9CQCh. 4 - Prob. 10CQCh. 4 - Prob. 11CQCh. 4 - Prob. 12CQCh. 4 - Prob. 13CQCh. 4 - Prob. 14CQCh. 4 - Prob. 15CQCh. 4 - Prob. 16CQCh. 4 - Prob. 17CQCh. 4 - Prob. 18CQCh. 4 - Prob. 19CQCh. 4 - Prob. 1MCQCh. 4 - Prob. 2MCQCh. 4 - Prob. 3MCQCh. 4 - Prob. 4MCQCh. 4 - Prob. 5MCQCh. 4 - Prob. 6MCQCh. 4 - Prob. 7MCQCh. 4 - Prob. 8MCQCh. 4 - Prob. 9MCQCh. 4 - Prob. 10MCQCh. 4 - Prob. 11MCQCh. 4 - Prob. 12MCQCh. 4 - Prob. 13MCQCh. 4 - Prob. 14MCQCh. 4 - Prob. 15MCQCh. 4 - Prob. 16MCQCh. 4 - Prob. 17MCQCh. 4 - Prob. 18MCQCh. 4 - Prob. 19MCQCh. 4 - Prob. 20MCQCh. 4 - Prob. 21MCQCh. 4 - Prob. 1PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19PCh. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Prob. 53PCh. 4 - Prob. 54PCh. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - Prob. 57PCh. 4 - Prob. 58PCh. 4 - Prob. 59PCh. 4 - Prob. 60PCh. 4 - Prob. 61PCh. 4 - Prob. 62PCh. 4 - Prob. 63PCh. 4 - Prob. 64PCh. 4 - Prob. 65PCh. 4 - Prob. 67PCh. 4 - Prob. 68PCh. 4 - Prob. 69PCh. 4 - Prob. 70PCh. 4 - Prob. 71PCh. 4 - Prob. 72PCh. 4 - Prob. 73PCh. 4 - Prob. 74PCh. 4 - Prob. 75PCh. 4 - Prob. 76PCh. 4 - Prob. 77PCh. 4 - Prob. 78PCh. 4 - Prob. 79PCh. 4 - Prob. 80PCh. 4 - Prob. 81PCh. 4 - Prob. 82PCh. 4 - Prob. 83PCh. 4 - Prob. 84PCh. 4 - Prob. 85PCh. 4 - Prob. 86PCh. 4 - Prob. 87PCh. 4 - Prob. 88PCh. 4 - Prob. 90PCh. 4 - Prob. 91PCh. 4 - Prob. 92PCh. 4 - Prob. 93PCh. 4 - Prob. 94PCh. 4 - Prob. 95PCh. 4 - Prob. 97PCh. 4 - Prob. 98PCh. 4 - Prob. 99PCh. 4 - Prob. 100PCh. 4 - Prob. 101PCh. 4 - Prob. 102PCh. 4 - Prob. 103PCh. 4 - Prob. 104PCh. 4 - Prob. 105P
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Kinematics Part 3: Projectile Motion; Author: Professor Dave explains;https://www.youtube.com/watch?v=aY8z2qO44WA;License: Standard YouTube License, CC-BY