Assessment Results » Golf Range
Individual
Report: Asha Hashim
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Total
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Hashim, Asha
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B
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C
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B
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A
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D
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(5/5)
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- The golf ball below has reached the top of its trajectory
in the time shown. What will the total hang time be when the ball hits the
ground? Assume no air resistance or spin.
- A. 6.15 seconds
- B. 8.20 seconds
- C. 12.30 seconds
- D. 16.40 seconds
Correct
Answer: B — 8.20 seconds
Explanation: Assuming no spin, the flight of a golf ball is symmetrical; it takes the same amount of time to go up as it takes to come down. Since it took 4.10 seconds to rise to the top of it's trajectory, it will take another 4.10 seconds to descend, for a total hang time of 8.20 seconds.
Explanation: Assuming no spin, the flight of a golf ball is symmetrical; it takes the same amount of time to go up as it takes to come down. Since it took 4.10 seconds to rise to the top of it's trajectory, it will take another 4.10 seconds to descend, for a total hang time of 8.20 seconds.
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- The golfer shown below has hit the ball with the
initial velocity and launch angle shown. What was the initial horizontal
velocity of the golf ball, vx?
- A. 29.0 m/s
- B. 38.7 m/s
- C. 41.0 m/s
- D. 45.0 m/s
Correct
Answer: C — 41.0 m/s
Explanation: To find the horizontal component of velocity, multiply the initial velocity by the cosine of the launch angle. The cosine of 45° is 0.7071, so the horizontal component of velocity is 58.0 • 0.7071 = 41.0 m/s.
Explanation: To find the horizontal component of velocity, multiply the initial velocity by the cosine of the launch angle. The cosine of 45° is 0.7071, so the horizontal component of velocity is 58.0 • 0.7071 = 41.0 m/s.
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- A cannon shoots a ball with an initial velocity of 152 m/s and a launch angle of 62°. The gravitational acceleration is 9.8 m/s2. Assuming no air resistance, how long will the cannonball stay in the air?
- A. 24.6 s
- B. 27.4 s
- C. 31.3 s
- D. 39.7 s
Correct
Answer: B — 27.4 s
Explanation: The hang time is independent of the horizontal velocity. To solve the problem, first calculate vertical velocity vector,
vy = vinitial • sin(?)
Using this, the cannonball's initial vertical velocity is
vy = 152 • sin(62) = 134.21 m/s
The next step is to calculate the hang time:
t = 2 • vy / g
t = 2 • 134.2 / 9.8
t = 27.4 s
Explanation: The hang time is independent of the horizontal velocity. To solve the problem, first calculate vertical velocity vector,
vy = vinitial • sin(?)
Using this, the cannonball's initial vertical velocity is
vy = 152 • sin(62) = 134.21 m/s
The next step is to calculate the hang time:
t = 2 • vy / g
t = 2 • 134.2 / 9.8
t = 27.4 s
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answered this question correctly!
- On February 6, 1971, Alan Shepard made history by becoming the first astronaut to play golf on the Moon. Using only one hand to swing the club because of his bulky space suit, Shepard still managed a long drive. Given the lack of an atmosphere on the moon and a gravitational acceleration of 1.6 m/s2, how far would the shot go with an initial velocity of 25 m/s and a launch angle of 35°?
- A. 367 m
- B. 375 m
- C. 382 m
- D. 391 m
Correct
Answer: A — 367 m
Explanation: To solve the problem, first calculate the horizontal and vertical velocity vectors,
vx = vinitial • cos(?)
vy = vinitial • sin(?)
Using this, the initial velocity vectors are
vx = 25 • cos(35) = 20.48 m/s
vy = 25 • sin(35) = 14.34 m/s
The next step is to calculate the hang time:
t = 2 • vy / g
t = 2 • 14.34 / 1.6
t = 17.92 s
Now, calculate the horizontal distance
d = vx • t
d = 20.48 • 17.92 = 367.1 m
Explanation: To solve the problem, first calculate the horizontal and vertical velocity vectors,
vx = vinitial • cos(?)
vy = vinitial • sin(?)
Using this, the initial velocity vectors are
vx = 25 • cos(35) = 20.48 m/s
vy = 25 • sin(35) = 14.34 m/s
The next step is to calculate the hang time:
t = 2 • vy / g
t = 2 • 14.34 / 1.6
t = 17.92 s
Now, calculate the horizontal distance
d = vx • t
d = 20.48 • 17.92 = 367.1 m
You
answered this question correctly!
- On the famous 18th hole at Pebble Beach, Lionel Forrest stands next to his ball on the fairway, 165 meters from the hole. His caddie hands him a golf club that will hit the ball at a 65° angle. The gravitational acceleration is 9.8 m/s2, and a stiff tailwind exactly cancels the effects of air resistance. How hard should Lionel hit the ball to land it as close to the hole as possible?
- A. 43 m/s
- B. 44 m/s
- C. 45 m/s
- D. 46 m/s
Correct
Answer: D — 46 m/s
Explanation: Lionel will land his ball closest to the hole if he hits it with an initial velocity of 46 m/s. The simplest method of solving the problem is by trial and error. To solve the problem using algebra, start with the equation for the distance the ball will travel,
d = vx • t
where t is the hang time of the ball. Also, we know that the hang time is twice the vertical velocity divided by gravitational acceleration,
t = 2 • vy / g
By substitution,
d = 2 vyvx / g
Finally, we know that
vx = v • cos(θ)
vy = v • sin(θ)
So
d = 2v2 cos(θ)sin(θ) / g
Solving for v2,
v2 = gd / 2cos(θ)sin(θ)
Substituting the given values for distance, gravity, and launch angle,
v2 = 9.8 • 165 / 2cos(65)sin(65)
v2 = 2,110.8
v = 45.9 m/s
Explanation: Lionel will land his ball closest to the hole if he hits it with an initial velocity of 46 m/s. The simplest method of solving the problem is by trial and error. To solve the problem using algebra, start with the equation for the distance the ball will travel,
d = vx • t
where t is the hang time of the ball. Also, we know that the hang time is twice the vertical velocity divided by gravitational acceleration,
t = 2 • vy / g
By substitution,
d = 2 vyvx / g
Finally, we know that
vx = v • cos(θ)
vy = v • sin(θ)
So
d = 2v2 cos(θ)sin(θ) / g
Solving for v2,
v2 = gd / 2cos(θ)sin(θ)
Substituting the given values for distance, gravity, and launch angle,
v2 = 9.8 • 165 / 2cos(65)sin(65)
v2 = 2,110.8
v = 45.9 m/s
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answered this question correctly!
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