A spacecraft(A) is launched from the surface of the Earth with a velocity of 0.8

Question

A spacecraft(A) is launched from the surface of the Earth with a velocity of 0.800c at an angle of 30.0° above the horizontal positive x-axis. Another spacecraft(B) is moving
past, with a velocity of 0.500c in the negative x-direction.
a) Determine the magnitude and direction of the velocity of the ?rst spacecraft as measured by the pilot of the second spacecraft.
b) What is the highest velocity in the negative x-direction for spacecraft B such that
relativistic effects contributes to less than 1% difference in the measured speed of
spacecraft(A)?

Explanation / Answer

Relative Y = (Uy + Vy)/(1 + UyVy/C^2) and relative X = (Ux + Vx)/(1 + UxVx/C^2) Uy = U sin(30) = .8C sin(30) = .4C; and Vy = 0 Ux = U cos(30) = .8C cos(30) = .693 C and Vx = .5 C So Y = (.4)C/(1 + .4*0) = .4 C and X = (1.193)C/(1 + .5*.693) = 0.886 C. a) Z = sqrt(.4^2 + .886^2) C = 0.972 C relative to each other at theta = ATAN(.4/.886) = 24.29940798 = 24.3 deg re the horizontal for A's direction to B. ANS. b) Sorry, I don't understand this question at all

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