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Let the circle be the running man.
Do you refer to the velocity of the rain raletaive to the ground, or relative to the stationary man, or relative to the running man? No matter what I will show all the cases.
vyvy is the velocity of the rain relative to the ground and to the stationary man, for we know the rain falls vertically if the man is at rest.
While the man is running, with reference to the running man, the magnitude of the horizontal component of the rain vxvx equals that of the running man such that the rain falls 60 degrees to the normal.
vx=12vx=12
vyvy remains unchanged with reference to the running man since the man only runs horizontally. In the triangle, vx=vytan60°vx=vytan60°.
Hence, we have
vytan60°=12vytan60°=12
vy=43–√vy=43 kmkm h−1h−1
Hence, the velocity of the rain relative to the ground and the stationary man is 43–√43 kmkm h−1h−1.
The velocity of the rain relative to the running man
=vx2+vy2−−−−−−−√=vx2+vy2
=122+(43–√)2−−−−−−−−−−−√=122+(43)2
=83–√=83 kmkm h−1h−1 (60° to the vertical line)
Do you refer to the velocity of the rain raletaive to the ground, or relative to the stationary man, or relative to the running man? No matter what I will show all the cases.
vyvy is the velocity of the rain relative to the ground and to the stationary man, for we know the rain falls vertically if the man is at rest.
While the man is running, with reference to the running man, the magnitude of the horizontal component of the rain vxvx equals that of the running man such that the rain falls 60 degrees to the normal.
vx=12vx=12
vyvy remains unchanged with reference to the running man since the man only runs horizontally. In the triangle, vx=vytan60°vx=vytan60°.
Hence, we have
vytan60°=12vytan60°=12
vy=43–√vy=43 kmkm h−1h−1
Hence, the velocity of the rain relative to the ground and the stationary man is 43–√43 kmkm h−1h−1.
The velocity of the rain relative to the running man
=vx2+vy2−−−−−−−√=vx2+vy2
=122+(43–√)2−−−−−−−−−−−√=122+(43)2
=83–√=83 kmkm h−1h−1 (60° to the vertical line)
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