Can you also help w C?C: Determine the direction of the velocity of the sandbag at a time 0.285 ss after its release.TIA A hot-air balloonist, rising vertically with a constant velocityof magnitude v = 5.00 m/s, releases a sandbag at aninstant when the balloon is a height h = 40.0 m above theground (Figure 1). After it is released, the sandbag is infree fall. For the questions that follow, take the origin of thecoordinate system used for measuring displacements to beat the ground, and upward displacements to be positive.Part ACompute the position of the sandbag at a time 0.285 s after its release.Express your answer in meters.?y =mSubmitRequest AnswerFigure1 of 1Part B0 - 500 m/sCompute the magnitude of the velocity of the sandbag at a time 0.285 s after its release.Express your answer in meters per second.?m/s40.0 m to groundSubmitRequest Answer

Name: Can you also help w C?C: Determine the direction of the velocity of th
Uploaded: 2024-03-05T11:56:34.000Z
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Description: Can you also help w C?C: Determine the direction of the velocity of the sandbag at a time 0.285 ss after its release.TIA A hot-air balloonist, rising vertically with a constant velocityof magnitude v = 5.00 m/s, releases a sandbag at aninstant when t

From the question, it is given that- Speed of balloon in air: vb = 5.00 m/s Initial speed of sandbag: vi = 0 m/s Initial distance of balloon from the ground: hi = 40.0 m Time taken: t1 = 0.285 s(a) From the second equation of kinematics, we have-Final distance of sandbag from ground:h=hi+vit+12at2Where, g is the acceleration due to gravity = -9.8 m/s2 (downward direction is negative)=>Final distance of sandbag from ground:h=40.0+0-9.82×(0.285)2 = 39.60 m The distance of the sandbag from the ground after 0.285 s is 39.60 m. (b) Let v be the speed of sandbag after time t = 0.285 s given by first equation of kinematics- v = u - gt On substituting the respected values in the above relation we get- v= -9.8 x 0.285 = -2.793 m/s. Speed of sandbag after t= 0.285 s is 2.793 m/s. (c) The direction of motion of sandbag is towards downwards as the speed of sandbag is -2.793 m/s where the negative sign shows downward direction. {Given upwards direction = positive.} Thus the direction of the velocity of sandbag after 0.285s of release is downwards.