A hot-air balloonist, rising vertically with a constant velocity of magnitude u

Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude u 5.00 m/s, releases a sandbag at an instant when the balloon is a height h 40.0 m above the ground (Figure 1). After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive Part C Compute the position of the sandbag at a time 1.00 s after its release Figure of 1 Submit quest Answer e 5.00 ms Part D Compute the velocity of the sandbag at a time 1.00 s after its release. u= m/s 40.0 m to ground Submit Request Answer

Explanation / Answer

C)

consider the motion in vertical direction

Yo = initial position of the sandbag at the time of release = h = 40 m

Y = final position of the sandbag after 1 s

a = acceleration = - 9.8 m/s2

Voy = initial velocity at the time of release = 5 m/s

t = time of travel = 1 sec

using the equation

Y = Yo + Voy t + (0.5) a t2

Y = 40 + 5 (1) + (0.5) (- 9.8) (1)2

Y = 40.1 m

D)

Vfy = final velocity

using the equation

Vfy = Voy + a t

Vfy = 5 + (- 9.8) (1)

Vfy = - 4.8 m/s

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