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A ball is thrown with an initial speed υi 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)
The time interval during which the ball is in motion.
Answer to Problem 4.56AP
The time interval during which the ball is in motion is √4R3g.
Explanation of Solution
Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi, the horizontal range of the ball is R and the maximum height covered by the ball is R6
The motion of the ball follows the parabolic path and the ball is said to projectile, the motion of the ball is shown in the Figure below.
Figure (1)
The formula to calculate the maximum height reached by the projectile is,
H=(vi)2sin2θ2g
Here,
g is the acceleration due to gravity.
H is the maximum height reached by the ball.
Rearrange the above equation.
(vi)2sin2θ=2gHvisinθ=√2gH
Substitute R6 for H in the above equation.
visinθ=√2gR6=√gR3
Thus, the vertical component of the initial velocity is √gR3.
The formula to calculate the time taken by the ball to reach the ground is,
t=2visinθg
Here,
t is the time taken by the ball to reach the ground.
Substitute √gR3 for visinθ in the above equation.
t=2√gR3g=√4R3g
Conclusion:
Therefore, the time interval during which the ball is in motion is √4R3g.
(b)
The speed of the ball at the peak of its path.
Answer to Problem 4.56AP
The speed of the ball at the peak of its path is √3gR4.
Explanation of Solution
Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi, the horizontal range of the ball is R and the maximum height covered by the ball is R6
From part (a) the time of flight is √4R3g.
From the Figure (1) the range of the ball and time of flight is,
R=(vicosθi)t
Rearrange the above equation.
vicosθi=Rt
Substitute √4R3g for t in the above equation.
vicosθi=R√4R3g=√3gR24R=√3gR4
Conclusion:
Therefore, the speed of the ball at the peak of its path is √3gR4.
(c)
The initial vertical component of the velocity.
Answer to Problem 4.56AP
The initial vertical component of the velocity is √gR3.
Explanation of Solution
Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi, the horizontal range of the ball is R and the maximum height covered by the ball is R6
From part (a) vertical component of the initial velocity is √gR3.
Conclusion:
Therefore, the initial vertical component of the velocity is √gR3.
(d)
The initial speed of the ball.
Answer to Problem 4.56AP
The initial velocity of the ball is √13gR12.
Explanation of Solution
Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi, the horizontal range of the ball is R and the maximum height covered by the ball is R6
From part (a) vertical component of the initial velocity is,
visinθi=√gR3
Square both side of the above equation.
(visinθi)2=(√gR3)2=gR3 (1)
And from part (b) the horizontal component of the velocity is √3gR4
vicosθi=√3gR4
Square both side of the above equation.
(vicosθi)2=(√3gR4)2=3gR4 (2)
Add equation (1) and (2) to find the initial velocity.
(visinθi)2+(vicosθi)2=3gR4+gR3(vi)2((sinθi)2+(cosθi)2)=9gR+4gR12(vi)2=13gR12vi=√13gR12
Conclusion:
Therefore, the initial velocity of the ball is √13gR12.
(e)
The angle θi.
Answer to Problem 4.56AP
The angle θi is 33.7°.
Explanation of Solution
Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi, the horizontal range of the ball is R and the maximum height covered by the ball is R6
From part (a) vertical component of the initial velocity is,
visinθi=√gR3
And from part (b) the horizontal component of the velocity is √3gR4
vicosθi=√3gR4
Take the ratio of the horizontal component and the vertical component of the initial velocity.
visinθivicosθi=√gR3√3gR4tanθi=√4gR9gRθi=tan−1√94=33.7°
Conclusion:
Therefore, the angle θi is 33.7°.
(f)
The maximum height that the ball can reach with the same initial velocity.
Answer to Problem 4.56AP
The maximum height that the ball can reach with the same initial velocity is 13R24.
Explanation of Solution
Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi, the horizontal range of the ball is R and the maximum height covered by the ball is R6
For the maximum height to be gained by the ball the angle made by the horizontal should be 90°.
The formula to calculate the maximum height reached by the projectile is,
Hmax=(vi)2sin290°2g
Here,
Hmax is the maximum height reached by the ball.
Rearrange the above equation.
Hmax=(vi)2sin290°2g=(vi)22g
From part (d) the initial velocity of the ball is,
vi=√13gR12
Substitute √13gR12 for vi in the above equation.
Hmax=(√13gR12)22g=13gR12(2g)=13R24
Conclusion:
Therefore, the maximum height that the ball can reach with the same initial velocity is 13R24.
(g)
The maximum range of the ball with the same initial velocity.
Answer to Problem 4.56AP
The maximum range of the ball with the same initial velocity is 13R12.
Explanation of Solution
Given info: The initial speed of the ball is vi and the angle made by the ball with the horizontal is θi, the horizontal range of the ball is R and the maximum height covered by the ball is R6
For the maximum range to be gained by the ball the angle made by the horizontal should be 45°.
The formula to calculate the maximum height reached by the projectile is,
Rmax=(vi)2sin22(45°)g=(vi)2g
Here,
Rmax is the maximum range of the ball.
From part (d) the initial velocity of the ball is,
vi=√13gR12
Substitute √13gR12 for vi in the above equation.
Rmax=(√13gR12)2g=13gR12(g)=13R12
Conclusion:
Therefore, the maximum range of the ball with the same initial velocity is 13R12.
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Chapter 4 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
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