An airplane with a speed of 97.5 m/s is climbing upward at anangle of 50degrees
ID: 1725363 • Letter: A
Question
An airplane with a speed of 97.5 m/s is climbing upward at anangle of 50degrees with respect to the horizontal. When the plane'saltitude is 732 m, the pilot drops a package. A. calculate the distance alongthe ground, measure beneath the points of release, to where thepackage hits the earth? B. Relative to the ground,determine the angle of the velocity vector of the package justbefore impact? An airplane with a speed of 97.5 m/s is climbing upward at anangle of 50degrees with respect to the horizontal. When the plane'saltitude is 732 m, the pilot drops a package. A. calculate the distance alongthe ground, measure beneath the points of release, to where thepackage hits the earth? B. Relative to the ground,determine the angle of the velocity vector of the package justbefore impact?Explanation / Answer
Initial speed of the package, U = 97.5 m/s Angle of projection, = 50o Altitude, h = 732 m Vertical displacement , Y = - 732m ( here '-' indicate downward direction ) Initial speed in horizontal direction, Ux = Ucos = 62.67 m/s Initial speed in Vertical direction , Uy = U sin = 74.7 m/s Vertical motion of the package: S = U t + ( 1/2 ) a t 2 Y = Uy t - ( 1/2 ) g t 2 - 732 = 74.7 t - 4.9 t 2 4.9 t2 - 74.7 t - 732 = 0 On solving the equation, time taken to reach theground, t = 22.02 s (a) Distance traveled along the ground , X = Uxt = 62.67 * 22.02 = 1380 m (b) Final velocity in horizontal direction, Vx = Ux= 62.67 m/s Final velocity in horizontal direction, Vy = Uy- g t = 74.7 - ( 9.8 * 22.02 ) = - 141.1 m/s Angle of velocity vector just before impact, = arc tan ( Vy / V x ) = arc tan ( - 141.1 / 62.67 ) = - 66o = 66o ( below the ground )
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