Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 4, Problem 4.65AP

A catapult launches a rocket at an angle of 53.0° above the horizontal with an initial speed of 100 m/s. The rocket engine immediately starts a burn, and for 3.00 s the rocket moves along its initial line of motion with an acceleration of 30.0 m/s2. Then its engine fails, and the rocket proceeds to move in free fall. Find (a) the maximum altitude reached by the rocket, (b) its total time of flight, and (c) its horizontal range.

(a)

Expert Solution
Check Mark
To determine

The maximum altitude reached by the rocket.

Answer to Problem 4.65AP

The maximum altitude reached by the rocket is 1.52km .

Explanation of Solution

Given info: The initial speed of the rocket is 100m/s with an angle 53.0° above the horizontal, the time for which the rocket runs along its initial line of motion is 3.00s and the acceleration is 30.0m/s2 .

The formula to calculate the vertical height reached by the rocket is,

h=((v0)ysinθ)t+12ayt2

Here,

h is the vertical height.

(v0)y is the vertical component of the initial velocity.

t is the time taken by the rocket.

ay is the vertical component acceleration of the rocket.

θ is the angle made by the rocket with the horizontal.

Substitute 100m/s for (v0)y , 53.0° for θ , 30.0m/s2 for ay and 3.00s for t in the above equation.

h=((100m/s)sin(53.0°))(3.00s)+12(30.0m/s2)(3.00s)2=((100m/s)sin(53.0°))(3.00s)+12(30.0m/s2)(3.00s)2=239.59m+135m=374.59m

Thus, the vertical height of rocket is 374.59m .

The speed of the rocket after the failure is,

v0=(v0)y+ayt

Substitute 100m/s for (v0)y , 30.0m/s2 for ay and 3.00s for t in the above equation.

v0=100m/s+(30.0m/s2)(3.00s)=190m/s

The height reached by the rocket after the engine failure is,

hf=v2v02sin2θ2g

Here,

v is the final velocity of the rocket.

g is the acceleration due to gravity.

hf is the height reached by the rocket after the engine failure.

Substitute 0 for v , 190m/s for v0 , 53.0° for θ and 10.0m/s2 for g in the above equation.

hf=02(190m/s)2sin2(53.0°)2(10.0m/s2)=1151.26m

The maximum height reached by the rocket is,

hmax=hf+h

Substitute 1151.26m for hf and 374.59m for h in the above equation.

hmax=1151.26m+374.59m=1525.8m(103km1m)=1.52km

Conclusion:

Therefore, the maximum altitude reached by the rocket is 1.52km .

(b)

Expert Solution
Check Mark
To determine

The total time of flight of the rocket.

Answer to Problem 4.65AP

The total time of flight of the rocket is 36.1s .

Explanation of Solution

Given info: The initial speed of the rocket is 100m/s with an angle 53.0° above the horizontal, the time for which the rocket runs along its initial line of motion is 3.00s and the acceleration is 30.0m/s2 .

The vertical height reached by the rocket at the time of free fall is,

h=(v0sinθ)t+12gt2

Here,

t is the time of flight of rocket after free fall.

Substitute 190m/s for v0 , 53.0° for θ , 9.8m/s2 for g and 374.59m for h in the above equation.

374.59m=(190m/s)sin(53.0°)t+12(9.8m/s2)t2374.59m=151.7m/st(4.9m/s2)t2t=33.10s,-2.14st=33.10s

The total time of the flight of the rocket is,

T=t+t

Substitute 33.10s for t and 3.00s for t in the above equation.

T=3.00s+33.10s=36.1s

Conclusion:

Therefore, the total time of flight of the rocket is 36.1s .

(c)

Expert Solution
Check Mark
To determine

The horizontal range of the rocket.

Answer to Problem 4.65AP

The horizontal range of the rocket is 4.05km .

Explanation of Solution

Given info: The initial speed of the rocket is 100m/s with an angle 53.0° above the horizontal, the time for which the rocket runs along its initial line of motion is 3.00s and the acceleration is 30.0m/s2 .

The formula to calculate the displacement of the rocket along the initial line of motion is,

h=(v0)yt+12ayt2

Here,

h is the displacement of the rocket along the initial line of motion.

Substitute 100m/s for (v0)y , 30.0m/s2 for ay and 3.00s for t in the above equation.

h=(100m/s)(3.00s)+12(30.0m/s2)(3.00s)2=300.0m+135.5m=435.0m

Thus, the displacement of the rocket along the initial line of motion is 435.0m .

The horizontal range of the rocket during constant acceleration is,

R=hcosθ

Here,

R is the horizontal range of the rocket during constant acceleration.

Substitute 435.0m for h and 53.0° for θ in the above equation.

R=(435.0m)cos(53.0°)=261.8m

The horizontal range of the rocket during free fall is calculated as,

R=(v0cosθ)T

Here,

R is the horizontal range of the rocket during free fall.

Substitute 190m/s for v0 , 53.0° for θ and 33.1s for T in the above equation.

R=(190m/s)cos(53.0°)(33.1s)=3784.81m

The total horizontal range of the rocket is,

R=R+R

Substitute 261.8m for R and 3784.81m for R in the above equation.

R=261.8m+3784.81m=4046.61m(103km1m)=4.0466km4.05km

Conclusion:

Therefore, the horizontal range of the rocket is 4.05km .

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Chapter 4 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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