An airplane with a speed of 74.9 m/s is climbing upward at an angle of 38.3 Solu

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i solved this using v=94.5m/s, angle=39.6degrees, altitude=713m. but the procedure is exactly same. hope you understand this. The vertical component of velocity of packet Uy = u x sinA* = 94.5 x sin 39.6 = 60.24 m/s The horizontal component of velocity of packet Ux = u x cosA* = 94.5 x cos 39.6 = 72.81 m/s Let the packet attain h meter due to Uy in t1 sec, =>By v^2 = u^2 - 2gh =>0 = (Uy)^2 - 2gh =>h = (60.24)^2/(2 x 9.8) =>h = 185.15 Thus the total height of the packet from the ground(H) = 713 + h = 713 + 185.15 = 898.15 m =>By v = u - gt =>0 = 60.24 - 9.8 x t1 =>t1 = 6.15 sec Let the packet take t2 sec to fall H, =>By s = ut + 1/2gt^2 =>898.15 = 0 + 1/2 x 9.8 x (t2)^2 =>t2 = sqrt[183.30] =>t2 = 13.54 sec Thus the total time of packet in air(T) = t1+t2 = 6.15 + 12.54 = 19.69 sec (a) Thus the horizontal distance covered by the packet = [Ux] x T = 72.81 x 19.69 = 1433.53 m (b) Let the final vertical velocity of the packet is Vy, =>By v^2 = u^2 + 2gh =>(Vy)^2 = 0 + 2 x 9.8 x 898.15 =>Vy = sqrt[17603.74] =>Vy = 132.68 m/s Thus the angle at which the packet will hit the ground(B*), By tanB* = Vy/Vx = 132/72.81 = 1.82 = ~61*2' =>B* = 61*2'

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