Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 3, Problem 54P

A ball is thrown with an initial speed vi at an angle θi with the horizontal. The horizontal range of the ball is R, and the ball reaches a maximum height R/6. In terms of R and g, find (a) the time interval during which the ball is in motion, (b) the ball’s speed at the peak of its path, (c) the initial vertical component of its velocity, (d) its initial speed, and (e) the angle θi. (f) Suppose the ball is thrown at the same initial speed found in (d) but at the angle appropriate for reaching the greatest height that it can. Find this height. (g) Suppose the ball is thrown at the same initial speed but at the angle for greatest possible range. Find this maximum horizontal range.

(a)

Expert Solution
Check Mark
To determine

Theflight time of the ball in the motion .

Answer to Problem 54P

The flight time of the ball in the motion is 2R3g_.

Explanation of Solution

Write the expression for the maximum height of the ball,

    h=vi2sin2θi2g        (I)

Here, h is the maximum height of the ball, vi is the initial speed of the ball, θi is the angle and g is the acceleration due to gravity.

Write the expression for the horizontal range of the ball,

    R=vi2sin22θig=2vi2sinθicosθig        (II)

Here, R is the horizontal range of the ball.

Substitute R6 for h in (I),

    visinθi=gR3        (III)

Combine (I) and (II) and substitute R6 for h and gR3 for visinθi,

    R=2(gR/3)vicosθig        (IV)

Rewrite the relation for vi.

    vicosθi=123gR        (V)

Write the expression for the vertical velocity of the ball,

    vyf=vyi+ayt        (VI)

Here, vyf is the final vertical velocity component of the ball, vyi is the initial vertical velocity component of the ball, ay is the vertical acceleration of the ball and t is the time taken.

Conclusion:

Substitute 0 for vyf, visinθi for vyi, tpeak for t and g for ay in (VI),

  0=visinθi+gtpeaktpeak=visinθi/gtpeak=gR3/g=R3g

Total time of the ball’s flight,

    tflight=2tpeak=2R3g

Therefore, theflight time of the ball in the motion is 2R3g_.

(b)

Expert Solution
Check Mark
To determine

Thespeed of the ball at the peak of its path.

Answer to Problem 54P

The speed of the ball at the peak of its path is vxi=123gR_.

Explanation of Solution

Write the expression for the speed of the ball at the path’s peak,

    vpeak=vxi=vicosθi        (VII)

Here, vpeak is the speed of the ball at the path’s peak, vxi is the initial horizontal velocity of the ball, vi is the initial speed of the ball and θi is the projectile angle of the ball.

Conclusion:

Substitute 123gR for vicosθi in the above equation,

    vpeak=123gR

Therefore, thespeed of the ball at the peak of its path is vxi=123gR_.

(c)

Expert Solution
Check Mark
To determine

The initial vertical component of the ball’s velocity .

Answer to Problem 54P

The initial vertical component of the ball’s velocity is vyi=gR3_.

Explanation of Solution

Write the expression for the initial vertical velocity of the ball,

    vyi=visinθi        (VIII)

Here, vyi is the initial vertical velocity of the ball, vi is the initial speed of the ball and θi is the projectile angle of the ball.

Conclusion:

Substitute gR3 for visinθi in (VIII),

    vyi=gR3

Therefore, theinitial vertical component of the ball’s velocity is vyi=gR3_.

(d)

Expert Solution
Check Mark
To determine

The initial speed of the ball.

Answer to Problem 54P

The initial speed of the ball is vi=13gR12_.

Explanation of Solution

Write the expression for the initial speed of the ball,

    vi2=vxi2+vyi2        (IX)

Here, vi is the initial speed of the ball, vyi is the initial vertical velocity of the ball, vxi is the initial horizontal velocity of the ball and θi is the projectile angle of the ball.

Rewrite the above expression,

    vi2=(visinθi)2+(vicosθi)2        (X)

Conclusion:

Substitute 123gR for vicosθi and gR3 for visinθi in (X),

    vi2=(gR3)2+(123gR)2=gR3+3gR4=13gR12

Therefore, theinitial speed of the ball is vi=13gR12_.

(e)

Expert Solution
Check Mark
To determine

The projectile angle of the ball .

Answer to Problem 54P

The projectile angle of the ballis θi=33.7o_.

Explanation of Solution

Equation (III) divided by (V),

    visinθivicosθi=[gR/3(123gR)]tanθi=23

Conclusion:

Rewrite the above equation,

    θi=tan1(23)=33.7o

Therefore, theprojectile angle of the ball is θi=33.7o_.

(f)

Expert Solution
Check Mark
To determine

The maximum height of the ball when it throws at maximum projection angle.

Answer to Problem 54P

The maximum height of the ball when it throws at maximum projection angle is 1324R_.

Explanation of Solution

In this case, the maximum projection angle must be 90.0o .

From (I),

Write the expression for the maximum height of the ball thrown,

    hmax=visin2θi2g

Here, hmax is the maximum height of the ball thrown.

Conclusion:

Substitute 1312gR for vi and 90.0o for θi in the above equation,

    hmax=(1312gR)sin290.0o2g=1324R

Therefore, themaximum height of the ball when it throws at maximum projection angle is 1324R_.

(g)

Expert Solution
Check Mark
To determine

The maximum horizontal range of the ball.

Answer to Problem 54P

The maximum horizontal range of the ballis 1312R_.

Explanation of Solution

In this case, the maximum projection angle must be 90.0o .

Write the expression for the maximum range of the ball thrown,

    Rmax=visinθig

Here, Rmax is the maximum range of the ball thrown.

Conclusion:

Substitute 1312gR for vi and 90.0o for θi in the above equation,

    Rmax=(1312gR)sin290.0og=1312R

Therefore, themaximum horizontal range of the ball is 1312R_.

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Chapter 3 Solutions

Principles of Physics: A Calculus-Based Text

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