According to Guinness the tallest man who ever lived was Robert. He was last measured in 1940 to be 2.72 (8 ft 11 in). Determine the speed in which a quarter would have reached before contact with the ground if dropped from rest from the top of his head.
Please help me answer this. I am confused
I have a problem I am stuck on?
2014-07-03 4:16 am
回答 (3)
2014-07-03 4:34 am
Initial velocity = u = 0
Final velocity = v
Distance = 2.72 m
gravitational acceleration = 9.8 m/s²
From one of the laws of motion we get
v²=u²+2as = 0 + 2(9.8)2.72 = 53.312
v = 7.3 m/s
Final velocity = v
Distance = 2.72 m
gravitational acceleration = 9.8 m/s²
From one of the laws of motion we get
v²=u²+2as = 0 + 2(9.8)2.72 = 53.312
v = 7.3 m/s
2014-07-03 4:28 am
t = √2h / g
v = at
plug the "t" equation into "v = at"
---
v = a√(2h / g)
v^2 = a^2/g * 2h
v^2 = 2ah
v = √(2ah)
---
We can do this in feet or meters, meters first:
m = 2.72
g = 9.81
v = √(2*9.81*2.72) = 7.3 m/s
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Feet
v = √(2 * 8.92 * 32.2) = 23.4 ft/s
v = at
plug the "t" equation into "v = at"
---
v = a√(2h / g)
v^2 = a^2/g * 2h
v^2 = 2ah
v = √(2ah)
---
We can do this in feet or meters, meters first:
m = 2.72
g = 9.81
v = √(2*9.81*2.72) = 7.3 m/s
---
Feet
v = √(2 * 8.92 * 32.2) = 23.4 ft/s
2014-07-03 4:39 am
Let s = the distance = 2.72 m
Let a = the acceleration due to gravity = 9.81 m/s²
Let v = the speed of the quarter
The equation for this can be derived from the equations for distance and speed
s = (1/2)(a)t²
v = (a)t
Substitute v/a for t into the first equation:
s = (1/2)(a)(v/a)²
s = (1/2)v²/a
2(a)s = v²
v = √{2(a)s}
Now you should be able to calculate the speed.
Let a = the acceleration due to gravity = 9.81 m/s²
Let v = the speed of the quarter
The equation for this can be derived from the equations for distance and speed
s = (1/2)(a)t²
v = (a)t
Substitute v/a for t into the first equation:
s = (1/2)(a)(v/a)²
s = (1/2)v²/a
2(a)s = v²
v = √{2(a)s}
Now you should be able to calculate the speed.
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