Asked by sam
A rocket of mass 4.36 × 105 kg is in flight near earth's surface. Its thrust is directed at an angle of 58.9° above the horizontal and has a magnitude of 9.61 × 106 N. Find the (a) magnitude and (b) direction of the rocket's acceleration. Give the direction as an angle above the horizontal.
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Answered by
Anonymous
F=ma, so a = F/m = 9.61*10^6 / 4.36*10^5 = 22.04 m/s^2
Take cos and sin of the angle to get the x- and y-components. Add -9.8 m/s^2 in the y-direction, and then find the magnitude and direction of the resultant.
Take cos and sin of the angle to get the x- and y-components. Add -9.8 m/s^2 in the y-direction, and then find the magnitude and direction of the resultant.
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