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Concept explainers
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)
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Theflight time of the ball in the motion .
Answer to Problem 54P
The flight time of the ball in the motion is 2√R3g_.
Explanation of Solution
Write the expression for the maximum height of the ball,
h=v2isin2θi2g (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=v2isin22θig=2v2isinθicosθig (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=12√3gR (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=2√R3g
Therefore, theflight time of the ball in the motion is 2√R3g_.
(b)
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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=12√3gR_.
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 12√3gR for vicosθi in the above equation,
vpeak=12√3gR
Therefore, thespeed of the ball at the peak of its path is vxi=12√3gR_.
(c)
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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)
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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,
v2i=v2xi+v2yi (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,
v2i=(visinθi)2+(vicosθi)2 (X)
Conclusion:
Substitute 12√3gR for vicosθi and √gR3 for visinθi in (X),
v2i=(√gR3)2+(12√3gR)2=gR3+3gR4=13gR12
Therefore, theinitial speed of the ball is vi=√13gR12_.
(e)
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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(12√3gR)]tanθi=23
Conclusion:
Rewrite the above equation,
θi=tan−1(23)=33.7o
Therefore, theprojectile angle of the ball is θi=33.7o_.
(f)
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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)
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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
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