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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 of7 m/s. Simultaneously an elevator cab starts to move upward from the same height the body was thrown from (the surface of the earth) witha 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 was2.45 m, and it took the body0.7 secto reach its max. height, and then I subtracted the height of the cab after 0.7 sec (= 2.1 m) from the2.45 m max heightand got0.35 mas the max height of the body relative to the cab's floor.

or2-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 of4 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 mor0.8 m) and why?

Thanks in advance!

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# Homework Help: Relative projectile motion

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