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NSTest5/MCAT Questions with
Detailed Verified Answers
If the average pitcher is releasing the ball from a height of 1.8 m above the
ground, and the pitcher's mound is 0.2 m higher than the rest of the baseball
field, at what height would the catcher need to hold his glove to caQuestion:
tch the pitched ball? (Note: neglect air resistance, estimate the acceleration
due to gravity as 10 m/s2, and assume the pitcher is only throwing the ball
horizontally.)
A. 2.0 m above the ground
B. 1.8 m above the ground
C. 0.5 m above the ground
D. 0.2 m above the ground
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Answer: d = rt for time
x = v(initial)t + 1/2at^2 (gravity for acceleration) to get distance ball falls
then subtract from initial height to get place catcher's glove needs to be
D is correct. The passage tells us that pitchers throw the ball at an average of
30 m/s and that the ball travels 18 m horizontally.
This means the ball's flight time is:
(18 m) / (30 m/s) = (18/30) s = 3/5 s = 0.6 s
The ball is released from a position 2 m off the ground (0.2 m from the
pitcher's mound and 1.8 m from the pitcher). To calculate the distance the ball
falls during 0.6 s, we can use the equation d = v0t + 1/2at2:
d = (0 m/s)(0.6 s) + 1/2(10 m/s2)(0.6 s)^2
d = 1/2(10)(0.36) = 1/2(3.6) = 1.8
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The ball has fallen 1.8 m from an initial height of 2.0 m. Thus, the catcher must
hold his glove 0.2 m above the ground to catch the pitch.
Question:
facultative anaerobe
Answer: makes ATP by aerobic
respiration if oxygen is present, but is
capable of switching to fermentation if
oxygen is absent
Question:
If the majority of the baseball's kinetic
energy comes from power generation in
the legs and hips, approximately how
much energy do the lower extremities
produce in the pitch?
, Page | 4
A. 65 J
B. 70 J
C. 140 J
D. 810 J
NSTest5/MCAT Questions with
Detailed Verified Answers
If the average pitcher is releasing the ball from a height of 1.8 m above the
ground, and the pitcher's mound is 0.2 m higher than the rest of the baseball
field, at what height would the catcher need to hold his glove to caQuestion:
tch the pitched ball? (Note: neglect air resistance, estimate the acceleration
due to gravity as 10 m/s2, and assume the pitcher is only throwing the ball
horizontally.)
A. 2.0 m above the ground
B. 1.8 m above the ground
C. 0.5 m above the ground
D. 0.2 m above the ground
, Page | 2
Answer: d = rt for time
x = v(initial)t + 1/2at^2 (gravity for acceleration) to get distance ball falls
then subtract from initial height to get place catcher's glove needs to be
D is correct. The passage tells us that pitchers throw the ball at an average of
30 m/s and that the ball travels 18 m horizontally.
This means the ball's flight time is:
(18 m) / (30 m/s) = (18/30) s = 3/5 s = 0.6 s
The ball is released from a position 2 m off the ground (0.2 m from the
pitcher's mound and 1.8 m from the pitcher). To calculate the distance the ball
falls during 0.6 s, we can use the equation d = v0t + 1/2at2:
d = (0 m/s)(0.6 s) + 1/2(10 m/s2)(0.6 s)^2
d = 1/2(10)(0.36) = 1/2(3.6) = 1.8
, Page | 3
The ball has fallen 1.8 m from an initial height of 2.0 m. Thus, the catcher must
hold his glove 0.2 m above the ground to catch the pitch.
Question:
facultative anaerobe
Answer: makes ATP by aerobic
respiration if oxygen is present, but is
capable of switching to fermentation if
oxygen is absent
Question:
If the majority of the baseball's kinetic
energy comes from power generation in
the legs and hips, approximately how
much energy do the lower extremities
produce in the pitch?
, Page | 4
A. 65 J
B. 70 J
C. 140 J
D. 810 J
