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A ball is projected horizontally with a speed 60 m/s from the top of the tall building the speed of the ball at t= 8s will be hey mate plz answer !!!! I need your help No spam!! |
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Answer» E!!! Since it is a horizontal projection therefore initially no vertical velocity is present. Things we know in x direction :Ux = 60 m/sax = 0Sx = 30t + 1/2(0)(t^2)SX = 60(8) = 480 mThings we know in y direction :Uy = 0m/say = gSy = (0)t + 1/2(10)(8^2)Sy = 320 mNow speed at t = 8 sec would be given by CONSIDERING both speed at t = 8 in y and x direction :Speed in x everytime remains constant, i. e., 60m/sSpeed in y at t = 8 is given by v = u +at ==> v = 0 + 10(8) = 80 m/sNow we consider both velocities as vectors so the velocity at t = 8 is :√(vx^2 + vy^2) => √(60^2 + 80^2 )=> √(3600 + 6400)=> √10000= 100 m/s Answer Hope it helps #Hakuna Matata :)) |
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