Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 4, Problem 73AP

A spring cannon is located at the edge of a table that is 1.20 m above the floor. A steel ball is launched from the cannon with speed vi at 35.0° above the horizontal. (a) Find the horizontal position of the ball as a function of vi at the instant it lands on the floor. We write this function as x(vi). Evaluate x for (b) vi = 0.100 m/s and for (c) vi = 100 m/s. (d) Assume vi is close to but not equal to zero. Show that one term in the answer to part (a) dominates so that the function x(vi) reduces to a simpler form. (c) If vi is very large, what is the approximate form of x(v)? (f) Describe the overall shape of the graph of the function x(vi).

(a)

Expert Solution
Check Mark
To determine

The horizontal position of the ball as a function of v1 .

Answer to Problem 73AP

The horizontal position of the ball as a function of v1 is v1(0.00229v12+0.1643)12+0.0479v12 .

Explanation of Solution

Given info: The located at the spring cannon is 1.2m above the floor and a ball is launched from the cannon with speed v1 at 35° above the horizontal.

Formula to calculate the vertical distance covered by the ball is,

y=yo+vyt12gt2 (1)

Here,

y is the vertical distance covered by the ball.

vy is the component of the velocity in y direction.

t is the time interval.

g is the acceleration due to gravity.

yo is the initial vertical distance above the floor.

The vertical component of the velocity is,

vy=v1sinθ

Here,

v1 is the velocity of the ball during projectile motion.

Substitute v1sinθ for vy in equation (1).

y=yo+v1sinθ×t12gt2

Substitute 1.2m for yo , 0 for y , 35.0° for θ and 9.81m/s2 for g to find t .

0=1.2m+v1sin35.0°×t129.81m/s2×t20=9.81m/s2×t22v1sin35.0°×t1.2m (2)

Solve the equation (2).

t=0.05846v1+(3.4184×103v12+0.2446)12

Formula to calculate the horizontal distance covered by the ball is,

x=vxt (3)

Here,

vx is the component of velocity in x direction.

t is the time interval.

The horizontal component of the velocity is,

vx=v1cosθ

Substitute v1cosθ for vx and 0.05846v1+(3.4184×103v12+0.2446)12 for t in equation (3).

x=v1cosθ×(0.05846v1+(3.4184×103v12+0.2446)12)

Substitute 35.0° for θ in above expression.

x=v1cos35.0°×(0.05846v1+(3.4184×103v12+0.2446)12)=v1(0.00229v12+0.1643)12+0.0479v12 (4)

Conclusion:

Therefore, the horizontal position of the ball as a function of v1 is v1(0.00229v12+0.1643)12+0.0479v12 .

(b)

Expert Solution
Check Mark
To determine

The horizontal position of the ball as v1=0.1m/s .

Answer to Problem 73AP

The horizontal position the ball as v1=0.1m/s . is 0.041m .

Explanation of Solution

Given info: The located at the spring cannon is 1.2m above the floor and a ball is launched from the cannon with speed v1 at 35° above the horizontal.

From equation (IV),

x=v1(0.00229v12+0.1643)12+0.0479v12

Substitute 0.1m/s for v1 in above expression.

x=0.1m/s(0.00229(0.1m/s)2+0.1643)12+0.0479(0.1m/s)2=0.041m

Conclusion:

Therefore, the horizontal position the ball as v1=0.1m/s . is 0.041m .

(c)

Expert Solution
Check Mark
To determine

The horizontal position of the ball as v1=100m/s .

Answer to Problem 73AP

The horizontal position the ball as v1=100m/s . is 960.18m .

Explanation of Solution

Given info: The located at the spring cannon is 1.2m above the floor and a ball is launched from the cannon with speed v1 at 35° above the horizontal.

From equation (4),

x=v1(0.00229v12+0.1643)12+0.0479v12

Substitute 100m/s for v1 in above expression.

x=100m/s(0.00229(100m/s)2+0.1643)12+0.0479(100m/s)2=960.18m

Conclusion:

Therefore, the horizontal position the ball as v1=0.1m/s . is 960.18m .

(d)

Expert Solution
Check Mark
To determine

The horizontal position of the ball as a function of v1 in a simpler form.

Answer to Problem 73AP

The horizontal position of the ball as a function of v1 in a simpler form is 0.405v1 .

Explanation of Solution

Given info: The located at the spring cannon is 1.2m above the floor and a ball is launched from the cannon with speed v1 at 35° above the horizontal.

From equation (IV),

x=v1(0.00229v12+0.1643)12+0.0479v12

The value of v1 is nearly to zero that is v120 .

Substitute 0 for v12 in above expression.

x=v1(0.00229×0+0.1643)12+0.0479×0=0.405v1

Conclusion:

Therefore, the horizontal position of the ball as a function of v1 in a simpler form is 0.405v1 .

(e)

Expert Solution
Check Mark
To determine

The horizontal position of the ball as a function of v1 if the value of v1 is very large.

Answer to Problem 73AP

The horizontal position of the ball as a function of v1 is 0.095v12 if the v1 is very large.

Explanation of Solution

Given info: The located at the spring cannon is 1.2m above the floor and a ball is launched from the cannon with speed v1 at 35° above the horizontal.

From equation (4),

x=v1(0.00229v12+0.1643)12+0.0479v12=(0.00229v14+0.1643v12)12+0.0479v12

The term is 0.1643v12 very small compare to the 0.00229v14 so the neglect the term 0.1643v12 .

x=(0.00229v14)12+0.0479v12=0.095v12

Conclusion:

Therefore, the horizontal position of the ball as a function of v1 is 0.095v12 if the v1 is very large.

(f)

Expert Solution
Check Mark
To determine

The overall shape of the graph of the function x(v) .

Answer to Problem 73AP

In starting condition graph x versus v1 is a straight line then increase the value of v1 curve become closer to the parabola.

Explanation of Solution

Given info: The located at the spring cannon is 1.2m above the floor and a ball is launched from the cannon with speed v1 at 35° above the horizontal.

The graph of x versus v1 start from the origin as a straight line with the slope is 0.405s then it curves upward above this tangent line, becoming a closer and closer to the parabola.

Conclusion:

Therefore, the starting condition graph x versus v1 is a straight line then increase the value of v1 curve become closer to the parabola.

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Chapter 4 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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