- #1
chem3
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Hello,
This is my first post here and I have the following question, please:
A body is thrown vertically upward from the surface of the Earth with an initial velocity of 7 m/s. Simultaneously an elevator cab starts to move upward from the same height the body was thrown from (the surface of the earth) with a constant velocity of 3 m/s. What is the max. height the body reaches relative to the cab's floor?
Two possible solutions I've thought of:
1- I calculated the max. height the body reaches relative to Earth's surface which was 2.45 m, and it took the body 0.7 sec to reach its max. height, and then I subtracted the height of the cab after 0.7 sec (= 2.1 m) from the 2.45 m max height and got 0.35 m as the max height of the body relative to the cab's floor.
or 2- I cosidered the cab as a frame of reference as it moves with a contant velocity, and that the body was thrown vertically with an initial velocity of 4 m/s (relative to the cab) and then used the equation:
V^2=Vi^2 + 2*g*H-max
with V=0 at max height and Vi=4 m/s (V of the body relative to the cab) and g=-10 m/s^2 we get:
H-max= max height = 16/20 = 0.8 m
Now, which solution is the right one (1 or 2 - 0.35 m or 0.8 m) and why?
Thanks in advance!
This is my first post here and I have the following question, please:
A body is thrown vertically upward from the surface of the Earth with an initial velocity of 7 m/s. Simultaneously an elevator cab starts to move upward from the same height the body was thrown from (the surface of the earth) with a constant velocity of 3 m/s. What is the max. height the body reaches relative to the cab's floor?
Two possible solutions I've thought of:
1- I calculated the max. height the body reaches relative to Earth's surface which was 2.45 m, and it took the body 0.7 sec to reach its max. height, and then I subtracted the height of the cab after 0.7 sec (= 2.1 m) from the 2.45 m max height and got 0.35 m as the max height of the body relative to the cab's floor.
or 2- I cosidered the cab as a frame of reference as it moves with a contant velocity, and that the body was thrown vertically with an initial velocity of 4 m/s (relative to the cab) and then used the equation:
V^2=Vi^2 + 2*g*H-max
with V=0 at max height and Vi=4 m/s (V of the body relative to the cab) and g=-10 m/s^2 we get:
H-max= max height = 16/20 = 0.8 m
Now, which solution is the right one (1 or 2 - 0.35 m or 0.8 m) and why?
Thanks in advance!
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