Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 3, Problem 69GP
To determine
The range of initial speeds allowed to make the basket.
Expert Solution & Answer
Answer to Problem 69GP
The minimum speed is 10.76 m/s and the maximum speed is 10.96 m/s .
Explanation of Solution
Given:
A basket ball leaves a player’s hand at a height of 2.10 m above the floor.
The distance between basket and floor is 2.60 m.
The player shoot the ball at a 38.0o angle.
A shot is made from the horizontal distance of 11.0 m with an accuracy of ±0.22 m .
Formula used:
Apply the second equation of the motion to determine the ranges of initial speeds. The formula is
s=ut+12at2
Where, s is the distance, u is the velocity, t is the time and a is the acceleration.
Calculation:
At a constant speed, the projectile will move with an acceleration, a=0 .
Le the player will travel a distance sd in time t .
The distance is
sd=11±0.22=11+0.22 and 11−0.22=11.22 m and 10.78 m
Minimum distance is
s1=10.78 m
Maximum distance is,
s2=11.22 m
When the motion is in horizontal direction, calculate distance using the formula,
sd=ut
sd=ucos(38°)tt=sducos(38°)
Figure 1
When the motion is in vertical direction, under the constant acceleration due to gravity, then the distance travelled by projectile is,
s=2.6−2.1s=0.5 m
Now, use the formula,
s=ut+12at2
Substitute the values,
0.5=usin(38°)sducos(38°)+12(−g)(sducos(38°))20.5=sdtan(38°)−gsd22u2cos2(38°)
Further solving the terms,
gsd22u2cos2(38°)=sdtan(38°)−0.5u2=gsd22cos2(38°)(sdtan(38°)−0.5)
Substitute 10.78 m for sd and 9.8 m/s2 for g to find the minimum velocity.
u2=115.867918u=10.76 m/s
Substitute 11.22 m for sd and 9.8 m/s2 for g to find the maximum velocity.
u2=120.29947u=10.96 m/s
Conclusion:
Therefore, the minimum range is 10.76 m/s and the maximum range is 10.96 m/s .
Chapter 3 Solutions
Physics: Principles with Applications
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